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bg-background","children":["$","$L3b",null,{"backHref":"/a-level/maths/ocr/questions","basePath":"/a-level/maths/ocr/integration-by-substitution/questions","formulaUrl":null,"initialIndex":36,"initialQuestionWindow":[{"id":"16baf928-c5e9-4af9-85de-ebb2b2243183","revisionId":"d157a28b-5e10-4b27-9ec5-4e6f92e33f53","questionNumber":null,"questionType":"LA","specification":"Perform the following integrations with respect to $x$:","difficulty":1,"difficultyBand":"easy","tier":null,"totalMarks":6,"sourceTitle":null,"calculatorUse":null,"isCalculation":false,"parts":[{"id":"333ba39b-c916-481e-879c-c07c403ec3cb","label":"a","content":"$$e^{5x-2}$","markscheme":"- Integrates exponential: $\\frac{1}{5}e^{5x-2} + C$ M1\n- Correct coefficient $\\frac{1}{5}$ A1\n- Includes constant of integration $+ C$ B1","marks":3,"order":0,"markingRubric":{"totalMarks":3,"points":[{"id":"p1","marks":1,"description":"Evidence of integrating to the form $ke^{5x-2}$.","code":"M"},{"id":"p2","marks":1,"description":"Correct coefficient of $\\frac{1}{5}$.","code":"A"},{"id":"p3","marks":1,"description":"Includes constant of integration $+ C$.","code":"B"}],"finalAnswers":["\\frac{1}{5}e^{5x-2} + C"],"generatedAt":"2026-05-06T14:04:17.306Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"direct_integration","label":"Direct Integration","steps":[{"type":"text","order":0,"title":"Recall the integration formula","content":"To integrate an exponential function of the form $e^{ax+b}$ with respect to $x$, we use the direct integration rule:\n\n$$\\int e^{ax+b} \\, dx = \\frac{1}{a}e^{ax+b} + C$$\n\nwhere $a$ is the coefficient of $x$ and $C$ is the constant of integration. \n\nThis rule works because of the reverse of the chain rule: when you differentiate $e^{ax+b}$, you multiply by the coefficient $a$. Therefore, when integrating, we must divide by $a$."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$e^{\\htmlData{anchor=coeff-a}{\\color{@sky}{5}}x \\htmlData{anchor=const-b}{\\color{@purple}{- 2}}}$"},{"latex":"$$\\htmlData{anchor=param-a}{\\color{@sky}{a}} = 5, \\quad \\htmlData{anchor=param-b}{\\color{@purple}{b}} = -2$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"coeff-a"},{"color":"purple","style":"text-color","anchor":"const-b"}],"connections":[{"to":"param-a","from":"coeff-a","color":"sky","label":"coefficient of x","style":"solid"},{"to":"param-b","from":"const-b","color":"purple","label":"constant term","style":"solid"}]},"order":1,"title":"Identify parameters","description":"By comparing the given integrand $e^{5x-2}$ with the general form $e^{ax+b}$, we can identify the coefficient of $x$ ($a$) and the constant term ($b$)."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\int e^{\\htmlData{anchor=source-exp}{\\color{@sky}{5}x-2}} \\, dx$"},{"latex":"$$= \\frac{1}{\\htmlData{anchor=target-coeff}{\\color{@sky}{5}}} e^{5x-2} + \\htmlData{anchor=const-c}{\\color{@pink}{C}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"source-exp"},{"color":"sky","style":"text-color","anchor":"target-coeff"},{"color":"pink","style":"text-color","anchor":"const-c"}],"connections":[{"to":"target-coeff","from":"source-exp","color":"sky","label":"divide by 5","style":"solid"}]},"order":2,"title":"Apply direct integration","description":"Now, we apply the formula by dividing the exponential term by the coefficient of $x$, which is $5$, and adding the constant of integration $C$."},{"type":"text","order":3,"title":"Final answer structure","content":"The completed integration gives:\n\n$$\\frac{1}{5}e^{5x-2} + C$$\n\n### Markscheme Breakdown\n- **M1**: Correctly integrating the exponential part to get an expression of the form $k e^{5x-2}$.\n- **A1**: Obtaining the correct coefficient of $\\frac{1}{5}$.\n- **B1**: Including the constant of integration $+ C$."}]},{"id":"substitution","label":"Integration by Substitution","steps":[{"type":"text","order":0,"title":"Set up the substitution","content":"We want to find the indefinite integral of the function with respect to $x$:\n\n$$\\int e^{5x-2} \\, dx$$\n\nTo simplify the integrand, we will use the formal integration by substitution method. Let's define our substitution variable $u$ as the exponent:\n\n$$u = 5x - 2$$"},{"type":"text","order":1,"title":"Find the differential dx","content":"Next, we find the relation between $du$ and $dx$ by differentiating $u$ with respect to $x$:\n\n$$\\frac{du}{dx} = 5$$\n\nTo substitute for $dx$ in our integral, we rearrange this equation to solve for $dx$:\n\n$$dx = \\frac{1}{5} du$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\int e^{\\htmlData{anchor=x-exp}{\\color{@sky}{5x-2}}} \\, \\htmlData{anchor=dx-term}{\\color{@amber}{dx}}$"},{"latex":"$$\\htmlData{anchor=u-term}{\\color{@sky}{u}} = 5x - 2 \\quad \\Rightarrow \\quad \\htmlData{anchor=dx-expr}{\\color{@amber}{dx}} = \\htmlData{anchor=du-expr}{\\color{@amber}{\\frac{1}{5} du}}$"},{"latex":"$$\\int e^{\\htmlData{anchor=u-sub}{\\color{@sky}{u}}} \\cdot \\htmlData{anchor=du-sub}{\\color{@amber}{\\frac{1}{5} du}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"x-exp"},{"color":"amber","style":"text-color","anchor":"dx-term"},{"color":"sky","style":"text-color","anchor":"u-term"},{"color":"amber","style":"text-color","anchor":"dx-expr"},{"color":"sky","style":"text-color","anchor":"u-sub"},{"color":"amber","style":"text-color","anchor":"du-sub"}],"connections":[{"to":"u-term","from":"x-exp","color":"sky","label":"substitute u","style":"solid"},{"to":"dx-expr","from":"dx-term","color":"amber","label":"substitute dx","style":"solid"},{"to":"u-sub","from":"u-term","color":"sky","label":null,"style":"solid"},{"to":"du-sub","from":"du-expr","color":"amber","label":null,"style":"solid"}]},"order":2,"title":"Substitute into the integral","description":"Now, we substitute both $u$ and $dx$ into the original integral to transform it entirely into a function of $u$."},{"type":"text","order":3,"title":"Integrate with respect to u","content":"We can factor out the constant coefficient $\\frac{1}{5}$ from the integral:\n\n$$\\frac{1}{5} \\int e^u \\, du$$\n\nUsing the standard integration rule $\\int e^u \\, du = e^u + C$, we perform the integration:\n\n$$\\frac{1}{5} e^u + C$$"},{"type":"text","order":4,"title":"Substitute back to find answer","content":"To write our final answer in terms of $x$, we substitute the original expression $u = 5x - 2$ back in:\n\n$$\\frac{1}{5}e^{5x-2} + C$$\n\n### Marks Distribution\n- **M1**: Integrating the exponential function to get the correct exponent form $k e^{5x-2}$\n- **A1**: Finding the correct coefficient of $\\frac{1}{5}$\n- **B1**: Including the constant of integration $+ C$"}]}],"generatedAt":"2026-06-09T03:29:40.651Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"dee81b26-f315-4e8f-90b4-4c46889010da","label":"b","content":"$$\\frac{1}{3 - 6x}$","markscheme":"- Integrates to natural log: $-\\frac{1}{6}\\ln|3-6x| + C$ M1\n- Correct coefficient $-\\frac{1}{6}$ A1\n- Includes constant of integration $+ C$ B1","marks":3,"order":1,"markingRubric":{"totalMarks":3,"points":[{"id":"p1","marks":1,"description":"Evidence of integrating to the form $k\\ln|3-6x|$.","code":"M"},{"id":"p2","marks":1,"description":"Correct coefficient of $-\\frac{1}{6}$.","code":"A"},{"id":"p3","marks":1,"description":"Includes constant of integration $+ C$.","code":"B"}],"finalAnswers":["-\\frac{1}{6}\\ln|3-6x| + C"],"generatedAt":"2026-05-06T14:04:17.306Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"direct","label":"Direct Integration","steps":[{"type":"text","order":0,"title":"Identify the standard rule","content":"To integrate a reciprocal function where the denominator is linear, we can apply the standard direct integration formula:\n\n$$\\int \\frac{1}{ax + b} \\, dx = \\frac{1}{a} \\ln|ax + b| + C$$\n\nHere, the integral is of the form $\\frac{1}{ax + b}$ where $a$ is the coefficient of $x$."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\int \\frac{1}{\\htmlData{anchor=const}{\\color{@sky}{3}} \\htmlData{anchor=coeff}{\\color{@amber}{- 6}}x} \\, dx$"},{"text":"Comparing this to the standard linear form gives:"},{"latex":"$$a = \\htmlData{anchor=a-val}{\\color{@amber}{-6}}, \\quad b = \\htmlData{anchor=b-val}{\\color{@sky}{3}}$"}],"highlights":[{"color":"amber","style":"box","anchor":"coeff"},{"color":"sky","style":"box","anchor":"const"}],"connections":[{"to":"a-val","from":"coeff","color":"amber","label":"coefficient of x","style":"solid"},{"to":"b-val","from":"const","color":"sky","label":"constant term","style":"solid"}]},"order":1,"title":"Identify coefficients","description":"We match our expression $\\frac{1}{3-6x}$ with the linear template $ax+b$ to find the values of $a$ and $b$."},{"type":"text","order":2,"title":"Apply direct integration","content":"Substitute $a = -6$ and $b = 3$ into the standard integration formula:\n\n$$\\int \\frac{1}{3 - 6x} \\, dx = \\frac{1}{-6} \\ln|3 - 6x| + C$$\n\nThis simplifies to:\n\n$$- \\frac{1}{6} \\ln|3 - 6x| + C$$"},{"type":"text","order":3,"title":"Review mark allocation","content":"According to the markscheme, the 3 marks are awarded as follows:\n- **M1**: Integrating to a natural log term of the form $k\\ln|3-6x|$.\n- **A1**: Obtaining the correct coefficient of $-\\frac{1}{6}$.\n- **B1**: Including the constant of integration $+ C$."}]},{"id":"substitution","label":"Integration by Substitution","steps":[{"type":"text","order":0,"title":"Set up the integral","content":"The question asks us to integrate the function $\\frac{1}{3 - 6x}$ with respect to $x$. We can write this as an indefinite integral:\n\n$$\\int \\frac{1}{3 - 6x} \\, dx$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$u = \\htmlData{anchor=u-expr}{\\color{@sky}{3 - 6x}}$"},{"latex":"$$\\frac{du}{dx} = \\htmlData{anchor=du-deriv}{\\color{@amber}{-6}}$"},{"latex":"$$dx = \\htmlData{anchor=dx-rearrange}{\\color{@purple}{-\\frac{1}{6}}} \\, du$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"u-expr"},{"color":"amber","style":"text-color","anchor":"du-deriv"},{"color":"purple","style":"text-color","anchor":"dx-rearrange"}],"connections":[{"to":"du-deriv","from":"u-expr","color":"amber","label":"differentiate","style":"solid"},{"to":"dx-rearrange","from":"du-deriv","color":"purple","label":"solve for dx","style":"solid"}]},"order":1,"title":"Define the substitution","description":"We choose a substitution variable $u$ to simplify the denominator. Differentiating $u$ with respect to $x$ allows us to express $dx$ in terms of $du$."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\int \\frac{1}{\\htmlData{anchor=denom-sub}{\\color{@sky}{u}}} \\, \\left(\\htmlData{anchor=dx-sub}{\\color{@purple}{-\\frac{1}{6}}} \\, du\\right)$"},{"latex":"$$\\htmlData{anchor=coeff-out}{\\color{@purple}{-\\frac{1}{6}}} \\int \\frac{1}{\\htmlData{anchor=u-int}{\\color{@sky}{u}}} \\, du$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"denom-sub"},{"color":"purple","style":"text-color","anchor":"dx-sub"},{"color":"purple","style":"text-color","anchor":"coeff-out"},{"color":"sky","style":"text-color","anchor":"u-int"}],"connections":[{"to":"coeff-out","from":"dx-sub","color":"purple","label":"factor out constant","style":"solid"}]},"order":2,"title":"Substitute into the integral","description":"Now, substitute $u = 3 - 6x$ and $dx = -\\frac{1}{6} \\, du$ into the integral. We then pull the constant factor $-\\frac{1}{6}$ outside of the integral."},{"type":"text","order":3,"title":"Integrate with respect to u","content":"Recall the standard integration formula from your formula booklet:\n\n$$\\int \\frac{1}{u} \\, du = \\ln|u| + C_0$$\n\nApplying this standard integral to our expression, we get:\n\n$$-\\frac{1}{6} \\int \\frac{1}{u} \\, du = -\\frac{1}{6} \\ln|u| + C$$\n\nwhere $C$ is the constant of integration."},{"type":"text","order":4,"title":"Substitute back to get x","content":"To write the final answer in terms of $x$, replace $u$ with our original substitution $u = 3 - 6x$:\n\n$$-\\frac{1}{6} \\ln|3 - 6x| + C$$\n\nThis completes the integration and matches the markscheme requirements: \n- Integrating to natural log form ($M1$)\n- Getting the correct coefficient $-\\frac{1}{6}$ ($A1$)\n- Including the $+ C$ constant ($B1$)"}]},{"id":"factoring","label":"Factoring the Denominator First","steps":[{"type":"text","order":0,"title":"Factor the denominator","content":"To integrate the expression using the factoring approach, we first look at the denominator of the integrand:\n\n$$3 - 6x$$\n\nWe can factor out a constant of $3$ from both terms in the denominator:\n\n$$3 - 6x = 3(1 - 2x)$$\n\nThis allows us to rewrite the integral as:\n\n$$\\int \\frac{1}{3 - 6x} \\, dx = \\int \\frac{1}{3(1 - 2x)} \\, dx$$"},{"type":"text","order":1,"title":"Factor out the constant","content":"Next, we can pull the constant factor of $\\frac{1}{3}$ outside the integral to make the integration simpler:\n\n$$\\int \\frac{1}{3(1 - 2x)} \\, dx = \\frac{1}{3} \\int \\frac{1}{1 - 2x} \\, dx$$"},{"type":"hyper_latex","block":{"nodes":[{"text":"Recall the standard rule: "},{"latex":"$$\\int \\frac{1}{ax + b} \\, dx = \\frac{1}{a} \\ln|ax + b| + C$"},{"latex":"$$\\text{Applying this with } \\htmlData{anchor=a-val}{a = -2} \\text{ gives:}$"},{"latex":"$$\\int \\frac{1}{1 - 2x} \\, dx = \\htmlData{anchor=coeff-res}{-\\frac{1}{2}} \\ln|1 - 2x| + C_1$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"a-val"},{"color":"sky","style":"text-color","anchor":"coeff-res"}],"connections":[{"to":"coeff-res","from":"a-val","color":"sky","label":"1/a coefficient","style":"solid"}]},"order":2,"title":"Integrate the function","description":"We now integrate the remaining function $\\frac{1}{1 - 2x}$ with respect to $x$ using the standard integration formula."},{"type":"text","order":3,"title":"Combine and simplify","content":"Now we multiply this result by the constant factor of $\\frac{1}{3}$ that we pulled outside the integral in Step 1:\n\n$$\\begin{aligned} \\int \\frac{1}{3 - 6x} \\, dx &= \\frac{1}{3} \\left( -\\frac{1}{2}\\ln|1 - 2x| \\right) + C \\\\ &= -\\frac{1}{6}\\ln|1 - 2x| + C \\end{aligned}$$\n\nSince $|3 - 6x| = 3|1 - 2x|$, this is mathematically equivalent to the markscheme's expression:\n\n$$- \\frac{1}{6}\\ln|3 - 6x| + C$$\n\nBoth forms are perfectly correct and will earn full marks!"}]}],"generatedAt":"2026-06-09T03:32:11.884Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"333ba39b-c916-481e-879c-c07c403ec3cb:direct_integration","label":"(a) Direct Integration","steps":[{"type":"text","order":0,"title":"Recall the integration formula","content":"To integrate an exponential function of the form $e^{ax+b}$ with respect to $x$, we use the direct integration rule:\n\n$$\\int e^{ax+b} \\, dx = \\frac{1}{a}e^{ax+b} + C$$\n\nwhere $a$ is the coefficient of $x$ and $C$ is the constant of integration. \n\nThis rule works because of the reverse of the chain rule: when you differentiate $e^{ax+b}$, you multiply by the coefficient $a$. Therefore, when integrating, we must divide by $a$."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$e^{\\htmlData{anchor=coeff-a}{\\color{@sky}{5}}x \\htmlData{anchor=const-b}{\\color{@purple}{- 2}}}$"},{"latex":"$$\\htmlData{anchor=param-a}{\\color{@sky}{a}} = 5, \\quad \\htmlData{anchor=param-b}{\\color{@purple}{b}} = -2$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"coeff-a"},{"color":"purple","style":"text-color","anchor":"const-b"}],"connections":[{"to":"param-a","from":"coeff-a","color":"sky","label":"coefficient of x","style":"solid"},{"to":"param-b","from":"const-b","color":"purple","label":"constant term","style":"solid"}]},"order":1,"title":"Identify parameters","description":"By comparing the given integrand $e^{5x-2}$ with the general form $e^{ax+b}$, we can identify the coefficient of $x$ ($a$) and the constant term ($b$)."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\int e^{\\htmlData{anchor=source-exp}{\\color{@sky}{5}x-2}} \\, dx$"},{"latex":"$$= \\frac{1}{\\htmlData{anchor=target-coeff}{\\color{@sky}{5}}} e^{5x-2} + \\htmlData{anchor=const-c}{\\color{@pink}{C}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"source-exp"},{"color":"sky","style":"text-color","anchor":"target-coeff"},{"color":"pink","style":"text-color","anchor":"const-c"}],"connections":[{"to":"target-coeff","from":"source-exp","color":"sky","label":"divide by 5","style":"solid"}]},"order":2,"title":"Apply direct integration","description":"Now, we apply the formula by dividing the exponential term by the coefficient of $x$, which is $5$, and adding the constant of integration $C$."},{"type":"text","order":3,"title":"Final answer structure","content":"The completed integration gives:\n\n$$\\frac{1}{5}e^{5x-2} + C$$\n\n### Markscheme Breakdown\n- **M1**: Correctly integrating the exponential part to get an expression of the form $k e^{5x-2}$.\n- **A1**: Obtaining the correct coefficient of $\\frac{1}{5}$.\n- **B1**: Including the constant of integration $+ C$."}]},{"id":"333ba39b-c916-481e-879c-c07c403ec3cb:substitution","label":"(a) Integration by Substitution","steps":[{"type":"text","order":0,"title":"Set up the substitution","content":"We want to find the indefinite integral of the function with respect to $x$:\n\n$$\\int e^{5x-2} \\, dx$$\n\nTo simplify the integrand, we will use the formal integration by substitution method. Let's define our substitution variable $u$ as the exponent:\n\n$$u = 5x - 2$$"},{"type":"text","order":1,"title":"Find the differential dx","content":"Next, we find the relation between $du$ and $dx$ by differentiating $u$ with respect to $x$:\n\n$$\\frac{du}{dx} = 5$$\n\nTo substitute for $dx$ in our integral, we rearrange this equation to solve for $dx$:\n\n$$dx = \\frac{1}{5} du$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\int e^{\\htmlData{anchor=x-exp}{\\color{@sky}{5x-2}}} \\, \\htmlData{anchor=dx-term}{\\color{@amber}{dx}}$"},{"latex":"$$\\htmlData{anchor=u-term}{\\color{@sky}{u}} = 5x - 2 \\quad \\Rightarrow \\quad \\htmlData{anchor=dx-expr}{\\color{@amber}{dx}} = \\htmlData{anchor=du-expr}{\\color{@amber}{\\frac{1}{5} du}}$"},{"latex":"$$\\int e^{\\htmlData{anchor=u-sub}{\\color{@sky}{u}}} \\cdot \\htmlData{anchor=du-sub}{\\color{@amber}{\\frac{1}{5} du}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"x-exp"},{"color":"amber","style":"text-color","anchor":"dx-term"},{"color":"sky","style":"text-color","anchor":"u-term"},{"color":"amber","style":"text-color","anchor":"dx-expr"},{"color":"sky","style":"text-color","anchor":"u-sub"},{"color":"amber","style":"text-color","anchor":"du-sub"}],"connections":[{"to":"u-term","from":"x-exp","color":"sky","label":"substitute u","style":"solid"},{"to":"dx-expr","from":"dx-term","color":"amber","label":"substitute dx","style":"solid"},{"to":"u-sub","from":"u-term","color":"sky","label":null,"style":"solid"},{"to":"du-sub","from":"du-expr","color":"amber","label":null,"style":"solid"}]},"order":2,"title":"Substitute into the integral","description":"Now, we substitute both $u$ and $dx$ into the original integral to transform it entirely into a function of $u$."},{"type":"text","order":3,"title":"Integrate with respect to u","content":"We can factor out the constant coefficient $\\frac{1}{5}$ from the integral:\n\n$$\\frac{1}{5} \\int e^u \\, du$$\n\nUsing the standard integration rule $\\int e^u \\, du = e^u + C$, we perform the integration:\n\n$$\\frac{1}{5} e^u + C$$"},{"type":"text","order":4,"title":"Substitute back to find answer","content":"To write our final answer in terms of $x$, we substitute the original expression $u = 5x - 2$ back in:\n\n$$\\frac{1}{5}e^{5x-2} + C$$\n\n### Marks Distribution\n- **M1**: Integrating the exponential function to get the correct exponent form $k e^{5x-2}$\n- **A1**: Finding the correct coefficient of $\\frac{1}{5}$\n- **B1**: Including the constant of integration $+ C$"}]},{"id":"dee81b26-f315-4e8f-90b4-4c46889010da:direct","label":"(b) Direct Integration","steps":[{"type":"text","order":0,"title":"Identify the standard rule","content":"To integrate a reciprocal function where the denominator is linear, we can apply the standard direct integration formula:\n\n$$\\int \\frac{1}{ax + b} \\, dx = \\frac{1}{a} \\ln|ax + b| + C$$\n\nHere, the integral is of the form $\\frac{1}{ax + b}$ where $a$ is the coefficient of $x$."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\int \\frac{1}{\\htmlData{anchor=const}{\\color{@sky}{3}} \\htmlData{anchor=coeff}{\\color{@amber}{- 6}}x} \\, dx$"},{"text":"Comparing this to the standard linear form gives:"},{"latex":"$$a = \\htmlData{anchor=a-val}{\\color{@amber}{-6}}, \\quad b = \\htmlData{anchor=b-val}{\\color{@sky}{3}}$"}],"highlights":[{"color":"amber","style":"box","anchor":"coeff"},{"color":"sky","style":"box","anchor":"const"}],"connections":[{"to":"a-val","from":"coeff","color":"amber","label":"coefficient of x","style":"solid"},{"to":"b-val","from":"const","color":"sky","label":"constant term","style":"solid"}]},"order":1,"title":"Identify coefficients","description":"We match our expression $\\frac{1}{3-6x}$ with the linear template $ax+b$ to find the values of $a$ and $b$."},{"type":"text","order":2,"title":"Apply direct integration","content":"Substitute $a = -6$ and $b = 3$ into the standard integration formula:\n\n$$\\int \\frac{1}{3 - 6x} \\, dx = \\frac{1}{-6} \\ln|3 - 6x| + C$$\n\nThis simplifies to:\n\n$$- \\frac{1}{6} \\ln|3 - 6x| + C$$"},{"type":"text","order":3,"title":"Review mark allocation","content":"According to the markscheme, the 3 marks are awarded as follows:\n- **M1**: Integrating to a natural log term of the form $k\\ln|3-6x|$.\n- **A1**: Obtaining the correct coefficient of $-\\frac{1}{6}$.\n- **B1**: Including the constant of integration $+ C$."}]},{"id":"dee81b26-f315-4e8f-90b4-4c46889010da:substitution","label":"(b) Integration by Substitution","steps":[{"type":"text","order":0,"title":"Set up the integral","content":"The question asks us to integrate the function $\\frac{1}{3 - 6x}$ with respect to $x$. We can write this as an indefinite integral:\n\n$$\\int \\frac{1}{3 - 6x} \\, dx$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$u = \\htmlData{anchor=u-expr}{\\color{@sky}{3 - 6x}}$"},{"latex":"$$\\frac{du}{dx} = \\htmlData{anchor=du-deriv}{\\color{@amber}{-6}}$"},{"latex":"$$dx = \\htmlData{anchor=dx-rearrange}{\\color{@purple}{-\\frac{1}{6}}} \\, du$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"u-expr"},{"color":"amber","style":"text-color","anchor":"du-deriv"},{"color":"purple","style":"text-color","anchor":"dx-rearrange"}],"connections":[{"to":"du-deriv","from":"u-expr","color":"amber","label":"differentiate","style":"solid"},{"to":"dx-rearrange","from":"du-deriv","color":"purple","label":"solve for dx","style":"solid"}]},"order":1,"title":"Define the substitution","description":"We choose a substitution variable $u$ to simplify the denominator. Differentiating $u$ with respect to $x$ allows us to express $dx$ in terms of $du$."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\int \\frac{1}{\\htmlData{anchor=denom-sub}{\\color{@sky}{u}}} \\, \\left(\\htmlData{anchor=dx-sub}{\\color{@purple}{-\\frac{1}{6}}} \\, du\\right)$"},{"latex":"$$\\htmlData{anchor=coeff-out}{\\color{@purple}{-\\frac{1}{6}}} \\int \\frac{1}{\\htmlData{anchor=u-int}{\\color{@sky}{u}}} \\, du$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"denom-sub"},{"color":"purple","style":"text-color","anchor":"dx-sub"},{"color":"purple","style":"text-color","anchor":"coeff-out"},{"color":"sky","style":"text-color","anchor":"u-int"}],"connections":[{"to":"coeff-out","from":"dx-sub","color":"purple","label":"factor out constant","style":"solid"}]},"order":2,"title":"Substitute into the integral","description":"Now, substitute $u = 3 - 6x$ and $dx = -\\frac{1}{6} \\, du$ into the integral. We then pull the constant factor $-\\frac{1}{6}$ outside of the integral."},{"type":"text","order":3,"title":"Integrate with respect to u","content":"Recall the standard integration formula from your formula booklet:\n\n$$\\int \\frac{1}{u} \\, du = \\ln|u| + C_0$$\n\nApplying this standard integral to our expression, we get:\n\n$$-\\frac{1}{6} \\int \\frac{1}{u} \\, du = -\\frac{1}{6} \\ln|u| + C$$\n\nwhere $C$ is the constant of integration."},{"type":"text","order":4,"title":"Substitute back to get x","content":"To write the final answer in terms of $x$, replace $u$ with our original substitution $u = 3 - 6x$:\n\n$$-\\frac{1}{6} \\ln|3 - 6x| + C$$\n\nThis completes the integration and matches the markscheme requirements: \n- Integrating to natural log form ($M1$)\n- Getting the correct coefficient $-\\frac{1}{6}$ ($A1$)\n- Including the $+ C$ constant ($B1$)"}]},{"id":"dee81b26-f315-4e8f-90b4-4c46889010da:factoring","label":"(b) Factoring the Denominator First","steps":[{"type":"text","order":0,"title":"Factor the denominator","content":"To integrate the expression using the factoring approach, we first look at the denominator of the integrand:\n\n$$3 - 6x$$\n\nWe can factor out a constant of $3$ from both terms in the denominator:\n\n$$3 - 6x = 3(1 - 2x)$$\n\nThis allows us to rewrite the integral as:\n\n$$\\int \\frac{1}{3 - 6x} \\, dx = \\int \\frac{1}{3(1 - 2x)} \\, dx$$"},{"type":"text","order":1,"title":"Factor out the constant","content":"Next, we can pull the constant factor of $\\frac{1}{3}$ outside the integral to make the integration simpler:\n\n$$\\int \\frac{1}{3(1 - 2x)} \\, dx = \\frac{1}{3} \\int \\frac{1}{1 - 2x} \\, dx$$"},{"type":"hyper_latex","block":{"nodes":[{"text":"Recall the standard rule: "},{"latex":"$$\\int \\frac{1}{ax + b} \\, dx = \\frac{1}{a} \\ln|ax + b| + C$"},{"latex":"$$\\text{Applying this with } \\htmlData{anchor=a-val}{a = -2} \\text{ gives:}$"},{"latex":"$$\\int \\frac{1}{1 - 2x} \\, dx = \\htmlData{anchor=coeff-res}{-\\frac{1}{2}} \\ln|1 - 2x| + C_1$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"a-val"},{"color":"sky","style":"text-color","anchor":"coeff-res"}],"connections":[{"to":"coeff-res","from":"a-val","color":"sky","label":"1/a coefficient","style":"solid"}]},"order":2,"title":"Integrate the function","description":"We now integrate the remaining function $\\frac{1}{1 - 2x}$ with respect to $x$ using the standard integration formula."},{"type":"text","order":3,"title":"Combine and simplify","content":"Now we multiply this result by the constant factor of $\\frac{1}{3}$ that we pulled outside the integral in Step 1:\n\n$$\\begin{aligned} \\int \\frac{1}{3 - 6x} \\, dx &= \\frac{1}{3} \\left( -\\frac{1}{2}\\ln|1 - 2x| \\right) + C \\\\ &= -\\frac{1}{6}\\ln|1 - 2x| + C \\end{aligned}$$\n\nSince $|3 - 6x| = 3|1 - 2x|$, this is mathematically equivalent to the markscheme's expression:\n\n$$- \\frac{1}{6}\\ln|3 - 6x| + C$$\n\nBoth forms are perfectly correct and will earn full marks!"}]}],"generatedAt":"2026-06-09T03:32:36.081Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null},{"id":"5a1c80f0-652d-4d35-a828-03ec3eace025","revisionId":"dd1516a4-4fe2-4abe-a535-6451d1114321","questionNumber":null,"questionType":"LA","specification":"","difficulty":1,"difficultyBand":"easy","tier":null,"totalMarks":5,"sourceTitle":null,"calculatorUse":null,"isCalculation":false,"parts":[{"id":"ffe89a9d-b428-473f-b05c-29a8c781c1d7","label":"a","content":"Use the substitution $u = \\tan x$ to find $\\int \\tan^5 x \\sec^2 x dx$","markscheme":"Let $u = \\tan x \\implies \\frac{du}{dx} = \\sec^2 x$ M1\nSubstitute: $\\int u^5 \\sec^2 x \\cdot \\frac{1}{\\sec^2 x} du = \\int u^5 du$ M1\nIntegrate: $\\frac{1}{6}u^6 + C$ M1\nSubstitute back: $\\frac{1}{6}\\tan^6 x + C$ A2","marks":5,"order":0,"markingRubric":{"totalMarks":5,"points":[{"id":"p1","marks":1,"description":"Differentiates substitution to find du/dx.","code":"M"},{"id":"p2","marks":1,"description":"Correctly substitutes into the integral to obtain a function of u only.","code":"M"},{"id":"p3","marks":1,"description":"Integrates the power of u correctly.","code":"M"},{"id":"p4","marks":2,"description":"Provides the final back-substituted expression with constant of integration.","code":"A","acceptable":["1/6 tan^6(x) + C","\\frac{\\tan^6 x}{6} + C"]}],"finalAnswers":["\\frac{1}{6}\\tan^6 x + C"],"generatedAt":"2026-05-06T14:04:28.557Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"formal_u_substitution","label":"Formal u-Substitution","steps":[{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{Let } \\htmlData{anchor=u-def}{\\color{@sky}{u = \\tan x}}$"},{"latex":"$$\\frac{du}{dx} = \\htmlData{anchor=du-deriv}{\\color{@amber}{\\sec^2 x}}$"},{"latex":"$$\\htmlData{anchor=dx-expr}{\\color{@purple}{dx = \\frac{du}{\\sec^2 x}}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"u-def"},{"color":"amber","style":"text-color","anchor":"du-deriv"}],"connections":[{"to":"du-deriv","from":"u-def","color":"amber","label":"differentiate","style":"solid"}]},"order":0,"title":"Define the substitution","description":"We start by setting $u = \\tan x$. Differentiating both sides with respect to $x$ gives $\\frac{du}{dx} = \\sec^2 x$, which we rearrange to express $dx$ in terms of $du$."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\int \\htmlData{anchor=orig-tan}{\\color{@sky}{\\tan^5 x}} \\sec^2 x \\, \\htmlData{anchor=orig-dx}{\\color{@purple}{dx}}$"},{"latex":"$$= \\int \\htmlData{anchor=sub-u}{\\color{@sky}{u^5}} \\sec^2 x \\cdot \\htmlData{anchor=sub-dx}{\\color{@purple}{\\frac{du}{\\sec^2 x}}}$"},{"latex":"$$= \\int u^5 \\, du$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"sub-u"},{"color":"purple","style":"text-color","anchor":"sub-dx"}],"connections":[{"to":"sub-u","from":"orig-tan","color":"sky","label":"u = tan x","style":"solid"},{"to":"sub-dx","from":"orig-dx","color":"purple","label":"substitute dx","style":"solid"}]},"order":1,"title":"Substitute into the integral","description":"Now, replace $\\tan x$ with $u$ and $dx$ with $\\frac{du}{\\sec^2 x}$. The $\\sec^2 x$ terms cancel out, leaving a simple integral in terms of $u$."},{"type":"text","order":2,"title":"Integrate with respect to u","content":"Using the power rule for integration, $\\int u^n \\, du = \\frac{u^{n+1}}{n+1} + C$, we integrate $u^5$:\n\n$$\\int u^5 \\, du = \\frac{1}{6}u^6 + C$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\frac{1}{6} \\htmlData{anchor=u-term}{\\color{@sky}{u}}^6 + C$"},{"latex":"$$= \\frac{1}{6} (\\htmlData{anchor=tan-term}{\\color{@sky}{\\tan x}})^6 + C$"},{"latex":"$$= \\frac{1}{6}\\tan^6 x + C$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"u-term"},{"color":"sky","style":"text-color","anchor":"tan-term"}],"connections":[{"to":"tan-term","from":"u-term","color":"sky","label":"substitute back","style":"solid"}]},"order":3,"title":"Substitute back to x","description":"Finally, replace $u$ with $\\tan x$ to write the final answer in terms of $x$. Don't forget to include the constant of integration, $C$."}]}],"generatedAt":"2026-06-09T05:13:36.924Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"ffe89a9d-b428-473f-b05c-29a8c781c1d7:formal_u_substitution","label":"Formal u-Substitution","steps":[{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{Let } \\htmlData{anchor=u-def}{\\color{@sky}{u = \\tan x}}$"},{"latex":"$$\\frac{du}{dx} = \\htmlData{anchor=du-deriv}{\\color{@amber}{\\sec^2 x}}$"},{"latex":"$$\\htmlData{anchor=dx-expr}{\\color{@purple}{dx = \\frac{du}{\\sec^2 x}}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"u-def"},{"color":"amber","style":"text-color","anchor":"du-deriv"}],"connections":[{"to":"du-deriv","from":"u-def","color":"amber","label":"differentiate","style":"solid"}]},"order":0,"title":"Define the substitution","description":"We start by setting $u = \\tan x$. Differentiating both sides with respect to $x$ gives $\\frac{du}{dx} = \\sec^2 x$, which we rearrange to express $dx$ in terms of $du$."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\int \\htmlData{anchor=orig-tan}{\\color{@sky}{\\tan^5 x}} \\sec^2 x \\, \\htmlData{anchor=orig-dx}{\\color{@purple}{dx}}$"},{"latex":"$$= \\int \\htmlData{anchor=sub-u}{\\color{@sky}{u^5}} \\sec^2 x \\cdot \\htmlData{anchor=sub-dx}{\\color{@purple}{\\frac{du}{\\sec^2 x}}}$"},{"latex":"$$= \\int u^5 \\, du$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"sub-u"},{"color":"purple","style":"text-color","anchor":"sub-dx"}],"connections":[{"to":"sub-u","from":"orig-tan","color":"sky","label":"u = tan x","style":"solid"},{"to":"sub-dx","from":"orig-dx","color":"purple","label":"substitute dx","style":"solid"}]},"order":1,"title":"Substitute into the integral","description":"Now, replace $\\tan x$ with $u$ and $dx$ with $\\frac{du}{\\sec^2 x}$. The $\\sec^2 x$ terms cancel out, leaving a simple integral in terms of $u$."},{"type":"text","order":2,"title":"Integrate with respect to u","content":"Using the power rule for integration, $\\int u^n \\, du = \\frac{u^{n+1}}{n+1} + C$, we integrate $u^5$:\n\n$$\\int u^5 \\, du = \\frac{1}{6}u^6 + C$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\frac{1}{6} \\htmlData{anchor=u-term}{\\color{@sky}{u}}^6 + C$"},{"latex":"$$= \\frac{1}{6} (\\htmlData{anchor=tan-term}{\\color{@sky}{\\tan x}})^6 + C$"},{"latex":"$$= \\frac{1}{6}\\tan^6 x + C$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"u-term"},{"color":"sky","style":"text-color","anchor":"tan-term"}],"connections":[{"to":"tan-term","from":"u-term","color":"sky","label":"substitute back","style":"solid"}]},"order":3,"title":"Substitute back to x","description":"Finally, replace $u$ with $\\tan x$ to write the final answer in terms of $x$. Don't forget to include the constant of integration, $C$."}]}],"generatedAt":"2026-06-09T05:13:47.925Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null},{"id":"d05dd575-82f2-4ddc-bfaf-fac22f730224","revisionId":"88a72f9f-8e39-48a6-a1be-925c1380a317","questionNumber":null,"questionType":"LA","specification":"Integrate with respect to $x$:","difficulty":1,"difficultyBand":"easy","tier":null,"totalMarks":6,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"bb36d3d3-454a-4d90-94e5-6791b703805f","label":"a","content":"$$\\sqrt{4x - 5}$","markscheme":"- Rewrites as $(4x-5)^{\\frac{1}{2}}$ and integrates to the form $k(4x-5)^{\\frac{3}{2}}$ M1\n- Simplifies to $\\frac{1}{6}(4x-5)^{\\frac{3}{2}}$ A1\n- Includes constant of integration $+ C$ B1","marks":3,"order":0,"markingRubric":{"totalMarks":3,"points":[{"id":"p1","marks":1,"description":"Attempt to integrate power to obtain $(4x-5)^{\\frac{3}{2}}$.","code":"M"},{"id":"p2","marks":1,"description":"Correct coefficient $\\frac{1}{6}$ or equivalent (from $\\frac{2}{3} \\times \\frac{1}{4}$).","code":"A"},{"id":"p3","marks":1,"description":"Include $+ C$ in the final answer.","code":"B"}],"finalAnswers":["\\frac{1}{6}(4x-5)^{\\frac{3}{2}} + C"],"generatedAt":"2026-05-06T14:04:11.791Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"reverse_chain_rule","label":"Reverse Chain Rule","steps":[{"type":"text","order":0,"title":"Rewrite as fractional power","content":"To integrate the square root function, we first rewrite the radical as a fractional exponent:\n\n$$\\int \\sqrt{4x - 5} \\, dx = \\int (4x - 5)^{\\frac{1}{2}} \\, dx$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$$\\int (\\htmlData{anchor=linear}{\\color{@sky}{4}}x - 5)^{\\htmlData{anchor=power}{\\color{@amber}{\\frac{1}{2}}}} \\, dx = \\frac{1}{\\htmlData{anchor=coeff}{\\color{@sky}{4}}} \\cdot \\frac{(4x - 5)^{\\htmlData{anchor=new-power}{\\color{@amber}{\\frac{3}{2}}}}}{\\htmlData{anchor=denom}{\\color{@amber}{\\frac{3}{2}}}} + C$$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"linear"},{"color":"sky","style":"text-color","anchor":"coeff"},{"color":"amber","style":"text-color","anchor":"power"},{"color":"amber","style":"text-color","anchor":"new-power"}],"connections":[{"to":"coeff","from":"linear","color":"sky","label":"divide by x-coefficient","style":"solid"},{"to":"new-power","from":"power","color":"amber","label":"add 1 to power","style":"solid"}]},"order":1,"title":"Apply reverse chain rule","description":"For any linear function inside a power, $\\int (ax+b)^n \\, dx = \\frac{1}{a} \\cdot \\frac{(ax+b)^{n+1}}{n+1} + C$. Here, the coefficient of $x$ is $4$, and the exponent increases from $\\frac{1}{2}$ to $\\frac{3}{2}$."},{"type":"text","order":2,"title":"Simplify the fraction","content":"Now, we simplify the coefficient in front of the term:\n\n$$\\frac{1}{4} \\cdot \\frac{1}{\\frac{3}{2}} = \\frac{1}{4} \\cdot \\frac{2}{3} = \\frac{2}{12} = \\frac{1}{6}$$\n\nThis gives us our final integrated form:\n\n$$\\frac{1}{6}(4x - 5)^{\\frac{3}{2}} + C$$"},{"type":"text","order":3,"title":"Mark scheme breakdown","content":"This question is worth **3 marks**:\n\n* **1 mark** (M1) for rewriting the expression in the form $(4x-5)^{\\frac{1}{2}}$ and performing integration to get $k(4x-5)^{\\frac{3}{2}}$\n* **1 mark** (A1) for simplifying the constant coefficient to $\\frac{1}{6}$\n* **1 mark** (B1) for including the constant of integration $+ C$"}]},{"id":"u_substitution","label":"Integration by Substitution","steps":[{"type":"text","order":0,"title":"Set up the integral","content":"We need to integrate the given expression with respect to $x$. Let's write this as an indefinite integral:\n\n$$\\int \\sqrt{4x - 5} \\, dx$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"u = \\htmlData{anchor=sub-u}{\\color{@sky}{4x - 5}}"},{"latex":"\\frac{du}{dx} = \\htmlData{anchor=diff-u}{\\color{@sky}{4}}"},{"latex":"dx = \\htmlData{anchor=sub-dx}{\\color{@pink}{\\frac{1}{4}}} \\, du"}],"highlights":[{"color":"sky","style":"underline","anchor":"sub-u"},{"color":"pink","style":"box","anchor":"sub-dx"}],"connections":[{"to":"diff-u","from":"sub-u","color":"sky","label":"differentiate","style":"solid"}]},"order":1,"title":"Define the substitution","description":"We choose $u = 4x - 5$ to simplify the expression inside the square root. Differentiating $u$ with respect to $x$ gives $\\frac{du}{dx} = 4$, which allows us to express $dx$ in terms of $du$."},{"type":"hyper_latex","block":{"nodes":[{"latex":"\\int \\sqrt{\\htmlData{anchor=orig-term}{\\color{@sky}{4x - 5}}} \\, \\htmlData{anchor=orig-dx}{\\color{@pink}{dx}}"},{"latex":"\\int \\sqrt{\\htmlData{anchor=new-term}{\\color{@sky}{u}}} \\cdot \\htmlData{anchor=new-dx}{\\color{@pink}{\\frac{1}{4} \\, du}}"},{"latex":"\\frac{1}{4} \\int u^{\\frac{1}{2}} \\, du"}],"highlights":[{"color":"sky","style":"text-color","anchor":"new-term"},{"color":"pink","style":"text-color","anchor":"new-dx"}],"connections":[{"to":"new-term","from":"orig-term","color":"sky","label":"substitute u","style":"solid"},{"to":"new-dx","from":"orig-dx","color":"pink","label":"substitute dx","style":"solid"}]},"order":2,"title":"Substitute into the integral","description":"Now we substitute both $u$ and $dx$ into our original integral. We can pull the constant factor of $\\frac{1}{4}$ outside the integral sign and rewrite the square root as a fractional power."},{"type":"text","order":3,"title":"Integrate with respect to u","content":"Using the power rule for integration, $\\int u^n \\, du = \\frac{u^{n+1}}{n+1}$ for $n = \\frac{1}{2}$:\n\n$$\\int u^{\\frac{1}{2}} \\, du = \\frac{u^{\\frac{3}{2}}}{\\frac{3}{2}} = \\frac{2}{3}u^{\\frac{3}{2}}$$\n\nMultiply this by our constant factor of $\\frac{1}{4}$:\n\n$$\\frac{1}{4} \\left( \\frac{2}{3}u^{\\frac{3}{2}} \\right) = \\frac{2}{12}u^{\\frac{3}{2}} = \\frac{1}{6}u^{\\frac{3}{2}}$$"},{"type":"text","order":4,"title":"Substitute back and add C","content":"Substitute $u = 4x - 5$ back into our integrated expression to get the final answer in terms of $x$, making sure to include the constant of integration $+ C$:\n\n$$\\frac{1}{6}(4x - 5)^{\\frac{3}{2}} + C$$"}]}],"generatedAt":"2026-06-11T03:46:11.180Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"9f953dfe-945e-4221-805a-2ed1006533fe","label":"b","content":"$$e^{6x + 1}$","markscheme":"- Integrates exponential to the form $k e^{6x+1}$ M1\n- Correct coefficient $\\frac{1}{6}$ A1\n- Includes constant of integration $+ C$ B1","marks":3,"order":1,"markingRubric":{"totalMarks":3,"points":[{"id":"p1","marks":1,"description":"Integrate to obtain $k e^{6x+1}$.","code":"M"},{"id":"p2","marks":1,"description":"Correct coefficient $\\frac{1}{6}$.","code":"A"},{"id":"p3","marks":1,"description":"Include $+ C$ in the final answer.","code":"B"}],"finalAnswers":["\\frac{1}{6}e^{6x+1} + C"],"generatedAt":"2026-05-06T14:04:11.791Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"direct_integration","label":"Direct Integration","steps":[{"type":"text","order":0,"title":"Identify the standard formula","content":"To integrate an exponential function of the form $e^{ax + b}$ directly, we use the standard integration formula:\n\n$$\\int e^{ax + b} \\, dx = \\frac{1}{a}e^{ax + b} + C$$\n\nwhere $a$ and $b$ are constants, and $C$ is the constant of integration."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$e^{\\htmlData{anchor=a-val}{\\color{@sky}{6}}x + \\htmlData{anchor=b-val}{\\color{@amber}{1}}}$"},{"latex":"$$a = \\color{@sky}{6}, \\quad b = \\color{@amber}{1}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"a-val"},{"color":"amber","style":"text-color","anchor":"b-val"}],"connections":[{"to":"b-val","from":"a-val","color":"sky","label":null,"style":"dashed"}]},"order":1,"title":"Match the parameters","description":"By comparing our expression $e^{6x + 1}$ to the general form $e^{ax+b}$, we can identify the constant values."},{"type":"text","order":2,"title":"Apply the formula","content":"Substituting $a = 6$ into our integration formula gives the coefficient $\\frac{1}{6}$:\n\n$$\\int e^{6x + 1} \\, dx = \\frac{1}{6}e^{6x + 1} + C$$"},{"type":"text","order":3,"title":"Understand the marks","content":"Let's review how marks are awarded for this question:\n\n* **M1**: Integrating the exponential function to the form $k e^{6x+1}$ (recognising that the exponential term remains unchanged).\n* **A1**: Getting the correct coefficient of $\\frac{1}{6}$.\n* **B1**: Including the constant of integration, $+ C$."}]},{"id":"substitution","label":"Integration by Substitution","steps":[{"type":"text","order":0,"title":"Define the integral","content":"We want to find the indefinite integral of the exponential function with respect to $x$:\n\n$$\\int e^{6x + 1} \\, dx$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$u = \\htmlData{anchor=u-expr}{6x + 1}$"},{"latex":"$$\\frac{du}{dx} = \\htmlData{anchor=deriv-val}{6}$"},{"latex":"$$dx = \\htmlData{anchor=dx-expr}{\\frac{1}{6}} \\, du$"}],"highlights":[{"color":"sky","style":"box","anchor":"u-expr"},{"color":"amber","style":"box","anchor":"dx-expr"}],"connections":[{"to":"deriv-val","from":"u-expr","color":"sky","label":"derivative","style":"solid"}]},"order":1,"title":"Set up the substitution","description":"Let's perform a substitution to simplify the exponent. We define $u$ as the linear expression in the exponent, then differentiate $u$ with respect to $x$ to find how the differentials $dx$ and $du$ relate."},{"type":"text","order":2,"title":"Substitute into the integral","content":"Now, substitute $u = 6x + 1$ and $dx = \\frac{1}{6} \\, du$ into the original integral:\n\n$$\\int e^{6x + 1} \\, dx = \\int e^u \\left(\\frac{1}{6} \\, du\\right)$$\n\nWe can pull the constant factor of $\\frac{1}{6}$ outside the integral:\n\n$$\\frac{1}{6} \\int e^u \\, du$$"},{"type":"text","order":3,"title":"Integrate with respect to u","content":"Using the standard integration rule for exponential functions, $\\int e^u \\, du = e^u$:\n\n$$\\frac{1}{6} \\int e^u \\, du = \\frac{1}{6} e^u + C$$\n\nwhere $C$ is the constant of integration."},{"type":"text","order":4,"title":"Substitute back to x","content":"Finally, replace $u$ with our original expression $6x + 1$ to write the final answer in terms of $x$:\n\n$$\\frac{1}{6} e^{6x + 1} + C$$"}]},{"id":"algebraic_simplification","label":"Algebraic Simplification First","steps":[{"type":"hyper_latex","block":{"nodes":[{"latex":"$$e^{\\htmlData{anchor=exp}{6x + 1}}$"},{"latex":"$$= \\htmlData{anchor=e-term}{\\color{@sky}{e^1}} \\cdot \\htmlData{anchor=ex-term}{\\color{@amber}{e^{6x}}}$"},{"latex":"$$= e \\cdot e^{6x}$"}],"highlights":[{"color":"sky","style":"box","anchor":"exp"}],"connections":[{"to":"e-term","from":"exp","color":"sky","label":"split indices","style":"dashed"}]},"order":0,"title":"Simplify using exponent rules","description":"We start by rewriting the integrand using the exponent rule $a^{m + n} = a^m \\cdot a^n$. This splits the constant and variable parts so we can treat the constant factor $e$ separately."},{"type":"text","order":1,"title":"Set up the integral","content":"Now we set up the integral of our simplified expression with respect to $x$:\n\n$$\\int e^{6x+1} \\, dx = \\int e \\cdot e^{6x} \\, dx$$\n\nBecause $e$ (or $e^1$) is a constant coefficient, we can pull it outside of the integration sign:\n\n$$e \\int e^{6x} \\, dx$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$e \\int \\htmlData{anchor=integrand}{\\color{@amber}{e^{6x}}} \\, dx$"},{"latex":"$$= e \\cdot \\left( \\htmlData{anchor=result}{\\color{@amber}{\\frac{1}{6} e^{6x}}} \\right) + C$"}],"highlights":[{"color":"amber","style":"text-color","anchor":"integrand"},{"color":"amber","style":"box","anchor":"result"}],"connections":[{"to":"result","from":"integrand","color":"amber","label":"integrate","style":"solid"}]},"order":2,"title":"Integrate the exponential term","description":"We integrate $e^{6x}$ using the standard integration rule $\\int e^{kx} \\, dx = \\frac{1}{k} e^{kx} + C$ where $k = 6$. This earns the first mark M1 for integrating to the form $k e^{6x+1}$ (or equivalent) and the second mark A1 for obtaining the correct coefficient $\\frac{1}{6}$."},{"type":"text","order":3,"title":"Recombine and state answer","content":"Now, multiply the constant $e$ back into the term and simplify using exponent rules:\n\n$$\\begin{aligned} \\frac{1}{6} e \\cdot e^{6x} + C &= \\frac{1}{6} e^1 \\cdot e^{6x} + C \\\\ &= \\frac{1}{6} e^{6x + 1} + C \\end{aligned}$$\n\nRemembering to add the constant of integration $+ C$ earns the final mark B1.\n\nOur final result is:\n\n$$\\frac{1}{6} e^{6x+1} + C$$"}]}],"generatedAt":"2026-06-11T03:46:24.788Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"bb36d3d3-454a-4d90-94e5-6791b703805f:reverse_chain_rule","label":"(a) Reverse Chain Rule","steps":[{"type":"text","order":0,"title":"Rewrite as fractional power","content":"To integrate the square root function, we first rewrite the radical as a fractional exponent:\n\n$$\\int \\sqrt{4x - 5} \\, dx = \\int (4x - 5)^{\\frac{1}{2}} \\, dx$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$$\\int (\\htmlData{anchor=linear}{\\color{@sky}{4}}x - 5)^{\\htmlData{anchor=power}{\\color{@amber}{\\frac{1}{2}}}} \\, dx = \\frac{1}{\\htmlData{anchor=coeff}{\\color{@sky}{4}}} \\cdot \\frac{(4x - 5)^{\\htmlData{anchor=new-power}{\\color{@amber}{\\frac{3}{2}}}}}{\\htmlData{anchor=denom}{\\color{@amber}{\\frac{3}{2}}}} + C$$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"linear"},{"color":"sky","style":"text-color","anchor":"coeff"},{"color":"amber","style":"text-color","anchor":"power"},{"color":"amber","style":"text-color","anchor":"new-power"}],"connections":[{"to":"coeff","from":"linear","color":"sky","label":"divide by x-coefficient","style":"solid"},{"to":"new-power","from":"power","color":"amber","label":"add 1 to power","style":"solid"}]},"order":1,"title":"Apply reverse chain rule","description":"For any linear function inside a power, $\\int (ax+b)^n \\, dx = \\frac{1}{a} \\cdot \\frac{(ax+b)^{n+1}}{n+1} + C$. Here, the coefficient of $x$ is $4$, and the exponent increases from $\\frac{1}{2}$ to $\\frac{3}{2}$."},{"type":"text","order":2,"title":"Simplify the fraction","content":"Now, we simplify the coefficient in front of the term:\n\n$$\\frac{1}{4} \\cdot \\frac{1}{\\frac{3}{2}} = \\frac{1}{4} \\cdot \\frac{2}{3} = \\frac{2}{12} = \\frac{1}{6}$$\n\nThis gives us our final integrated form:\n\n$$\\frac{1}{6}(4x - 5)^{\\frac{3}{2}} + C$$"},{"type":"text","order":3,"title":"Mark scheme breakdown","content":"This question is worth **3 marks**:\n\n* **1 mark** (M1) for rewriting the expression in the form $(4x-5)^{\\frac{1}{2}}$ and performing integration to get $k(4x-5)^{\\frac{3}{2}}$\n* **1 mark** (A1) for simplifying the constant coefficient to $\\frac{1}{6}$\n* **1 mark** (B1) for including the constant of integration $+ C$"}]},{"id":"bb36d3d3-454a-4d90-94e5-6791b703805f:u_substitution","label":"(a) Integration by Substitution","steps":[{"type":"text","order":0,"title":"Set up the integral","content":"We need to integrate the given expression with respect to $x$. Let's write this as an indefinite integral:\n\n$$\\int \\sqrt{4x - 5} \\, dx$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"u = \\htmlData{anchor=sub-u}{\\color{@sky}{4x - 5}}"},{"latex":"\\frac{du}{dx} = \\htmlData{anchor=diff-u}{\\color{@sky}{4}}"},{"latex":"dx = \\htmlData{anchor=sub-dx}{\\color{@pink}{\\frac{1}{4}}} \\, du"}],"highlights":[{"color":"sky","style":"underline","anchor":"sub-u"},{"color":"pink","style":"box","anchor":"sub-dx"}],"connections":[{"to":"diff-u","from":"sub-u","color":"sky","label":"differentiate","style":"solid"}]},"order":1,"title":"Define the substitution","description":"We choose $u = 4x - 5$ to simplify the expression inside the square root. Differentiating $u$ with respect to $x$ gives $\\frac{du}{dx} = 4$, which allows us to express $dx$ in terms of $du$."},{"type":"hyper_latex","block":{"nodes":[{"latex":"\\int \\sqrt{\\htmlData{anchor=orig-term}{\\color{@sky}{4x - 5}}} \\, \\htmlData{anchor=orig-dx}{\\color{@pink}{dx}}"},{"latex":"\\int \\sqrt{\\htmlData{anchor=new-term}{\\color{@sky}{u}}} \\cdot \\htmlData{anchor=new-dx}{\\color{@pink}{\\frac{1}{4} \\, du}}"},{"latex":"\\frac{1}{4} \\int u^{\\frac{1}{2}} \\, du"}],"highlights":[{"color":"sky","style":"text-color","anchor":"new-term"},{"color":"pink","style":"text-color","anchor":"new-dx"}],"connections":[{"to":"new-term","from":"orig-term","color":"sky","label":"substitute u","style":"solid"},{"to":"new-dx","from":"orig-dx","color":"pink","label":"substitute dx","style":"solid"}]},"order":2,"title":"Substitute into the integral","description":"Now we substitute both $u$ and $dx$ into our original integral. We can pull the constant factor of $\\frac{1}{4}$ outside the integral sign and rewrite the square root as a fractional power."},{"type":"text","order":3,"title":"Integrate with respect to u","content":"Using the power rule for integration, $\\int u^n \\, du = \\frac{u^{n+1}}{n+1}$ for $n = \\frac{1}{2}$:\n\n$$\\int u^{\\frac{1}{2}} \\, du = \\frac{u^{\\frac{3}{2}}}{\\frac{3}{2}} = \\frac{2}{3}u^{\\frac{3}{2}}$$\n\nMultiply this by our constant factor of $\\frac{1}{4}$:\n\n$$\\frac{1}{4} \\left( \\frac{2}{3}u^{\\frac{3}{2}} \\right) = \\frac{2}{12}u^{\\frac{3}{2}} = \\frac{1}{6}u^{\\frac{3}{2}}$$"},{"type":"text","order":4,"title":"Substitute back and add C","content":"Substitute $u = 4x - 5$ back into our integrated expression to get the final answer in terms of $x$, making sure to include the constant of integration $+ C$:\n\n$$\\frac{1}{6}(4x - 5)^{\\frac{3}{2}} + C$$"}]},{"id":"9f953dfe-945e-4221-805a-2ed1006533fe:direct_integration","label":"(b) Direct Integration","steps":[{"type":"text","order":0,"title":"Identify the standard formula","content":"To integrate an exponential function of the form $e^{ax + b}$ directly, we use the standard integration formula:\n\n$$\\int e^{ax + b} \\, dx = \\frac{1}{a}e^{ax + b} + C$$\n\nwhere $a$ and $b$ are constants, and $C$ is the constant of integration."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$e^{\\htmlData{anchor=a-val}{\\color{@sky}{6}}x + \\htmlData{anchor=b-val}{\\color{@amber}{1}}}$"},{"latex":"$$a = \\color{@sky}{6}, \\quad b = \\color{@amber}{1}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"a-val"},{"color":"amber","style":"text-color","anchor":"b-val"}],"connections":[{"to":"b-val","from":"a-val","color":"sky","label":null,"style":"dashed"}]},"order":1,"title":"Match the parameters","description":"By comparing our expression $e^{6x + 1}$ to the general form $e^{ax+b}$, we can identify the constant values."},{"type":"text","order":2,"title":"Apply the formula","content":"Substituting $a = 6$ into our integration formula gives the coefficient $\\frac{1}{6}$:\n\n$$\\int e^{6x + 1} \\, dx = \\frac{1}{6}e^{6x + 1} + C$$"},{"type":"text","order":3,"title":"Understand the marks","content":"Let's review how marks are awarded for this question:\n\n* **M1**: Integrating the exponential function to the form $k e^{6x+1}$ (recognising that the exponential term remains unchanged).\n* **A1**: Getting the correct coefficient of $\\frac{1}{6}$.\n* **B1**: Including the constant of integration, $+ C$."}]},{"id":"9f953dfe-945e-4221-805a-2ed1006533fe:substitution","label":"(b) Integration by Substitution","steps":[{"type":"text","order":0,"title":"Define the integral","content":"We want to find the indefinite integral of the exponential function with respect to $x$:\n\n$$\\int e^{6x + 1} \\, dx$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$u = \\htmlData{anchor=u-expr}{6x + 1}$"},{"latex":"$$\\frac{du}{dx} = \\htmlData{anchor=deriv-val}{6}$"},{"latex":"$$dx = \\htmlData{anchor=dx-expr}{\\frac{1}{6}} \\, du$"}],"highlights":[{"color":"sky","style":"box","anchor":"u-expr"},{"color":"amber","style":"box","anchor":"dx-expr"}],"connections":[{"to":"deriv-val","from":"u-expr","color":"sky","label":"derivative","style":"solid"}]},"order":1,"title":"Set up the substitution","description":"Let's perform a substitution to simplify the exponent. We define $u$ as the linear expression in the exponent, then differentiate $u$ with respect to $x$ to find how the differentials $dx$ and $du$ relate."},{"type":"text","order":2,"title":"Substitute into the integral","content":"Now, substitute $u = 6x + 1$ and $dx = \\frac{1}{6} \\, du$ into the original integral:\n\n$$\\int e^{6x + 1} \\, dx = \\int e^u \\left(\\frac{1}{6} \\, du\\right)$$\n\nWe can pull the constant factor of $\\frac{1}{6}$ outside the integral:\n\n$$\\frac{1}{6} \\int e^u \\, du$$"},{"type":"text","order":3,"title":"Integrate with respect to u","content":"Using the standard integration rule for exponential functions, $\\int e^u \\, du = e^u$:\n\n$$\\frac{1}{6} \\int e^u \\, du = \\frac{1}{6} e^u + C$$\n\nwhere $C$ is the constant of integration."},{"type":"text","order":4,"title":"Substitute back to x","content":"Finally, replace $u$ with our original expression $6x + 1$ to write the final answer in terms of $x$:\n\n$$\\frac{1}{6} e^{6x + 1} + C$$"}]},{"id":"9f953dfe-945e-4221-805a-2ed1006533fe:algebraic_simplification","label":"(b) Algebraic Simplification First","steps":[{"type":"hyper_latex","block":{"nodes":[{"latex":"$$e^{\\htmlData{anchor=exp}{6x + 1}}$"},{"latex":"$$= \\htmlData{anchor=e-term}{\\color{@sky}{e^1}} \\cdot \\htmlData{anchor=ex-term}{\\color{@amber}{e^{6x}}}$"},{"latex":"$$= e \\cdot e^{6x}$"}],"highlights":[{"color":"sky","style":"box","anchor":"exp"}],"connections":[{"to":"e-term","from":"exp","color":"sky","label":"split indices","style":"dashed"}]},"order":0,"title":"Simplify using exponent rules","description":"We start by rewriting the integrand using the exponent rule $a^{m + n} = a^m \\cdot a^n$. This splits the constant and variable parts so we can treat the constant factor $e$ separately."},{"type":"text","order":1,"title":"Set up the integral","content":"Now we set up the integral of our simplified expression with respect to $x$:\n\n$$\\int e^{6x+1} \\, dx = \\int e \\cdot e^{6x} \\, dx$$\n\nBecause $e$ (or $e^1$) is a constant coefficient, we can pull it outside of the integration sign:\n\n$$e \\int e^{6x} \\, dx$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$e \\int \\htmlData{anchor=integrand}{\\color{@amber}{e^{6x}}} \\, dx$"},{"latex":"$$= e \\cdot \\left( \\htmlData{anchor=result}{\\color{@amber}{\\frac{1}{6} e^{6x}}} \\right) + C$"}],"highlights":[{"color":"amber","style":"text-color","anchor":"integrand"},{"color":"amber","style":"box","anchor":"result"}],"connections":[{"to":"result","from":"integrand","color":"amber","label":"integrate","style":"solid"}]},"order":2,"title":"Integrate the exponential term","description":"We integrate $e^{6x}$ using the standard integration rule $\\int e^{kx} \\, dx = \\frac{1}{k} e^{kx} + C$ where $k = 6$. This earns the first mark M1 for integrating to the form $k e^{6x+1}$ (or equivalent) and the second mark A1 for obtaining the correct coefficient $\\frac{1}{6}$."},{"type":"text","order":3,"title":"Recombine and state answer","content":"Now, multiply the constant $e$ back into the term and simplify using exponent rules:\n\n$$\\begin{aligned} \\frac{1}{6} e \\cdot e^{6x} + C &= \\frac{1}{6} e^1 \\cdot e^{6x} + C \\\\ &= \\frac{1}{6} e^{6x + 1} + C \\end{aligned}$$\n\nRemembering to add the constant of integration $+ C$ earns the final mark B1.\n\nOur final result is:\n\n$$\\frac{1}{6} e^{6x+1} + C$$"}]}],"generatedAt":"2026-06-11T03:46:24.836Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null}],"subjectCode":"a_level_maths__ocr","summaries":[{"id":"04b8b144-d788-4727-9d45-7e3c324a4881","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":false},{"id":"41fb9ee5-982a-4e7b-ba09-a06e0a62b120","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"4a070e1e-0c19-4c8f-b5c8-8e74f42e2cb0","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":false},{"id":"64c0784b-74d4-4c3b-ad01-85bb57eca6f8","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":false},{"id":"678b8a53-192a-4dc8-8642-f373e1ea4601","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"6a3c5f33-7ee6-4203-9e71-d6f699fe93be","difficultyBand":"easy","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"7242d354-9652-4ce4-b93c-a86a27c4f362","difficultyBand":"easy","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":false},{"id":"88df656e-aa1a-4c3c-97df-21ce440984ae","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":false},{"id":"90d5eaaf-c081-4172-9fb6-0bb687060924","difficultyBand":"medium","calculatorUse":null,"totalMarks":7,"questionType":"LA","isCalculation":false},{"id":"9d98ac71-67b3-4622-a4f2-a035d1537bdf","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":false},{"id":"c4e6b947-384c-4f92-8ef3-0ef73e7dafb0","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"cd4a2b44-8258-4eac-a7e7-ba92c0baf455","difficultyBand":"medium","calculatorUse":null,"totalMarks":7,"questionType":"LA","isCalculation":true},{"id":"cf8a0e30-159c-4ad3-aeee-9692e1ae78ae","difficultyBand":"easy","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":false},{"id":"ded7644d-5c2f-4726-b6a2-47c930d80aec","difficultyBand":"medium","calculatorUse":null,"totalMarks":7,"questionType":"LA","isCalculation":false},{"id":"1c062306-1388-4be0-af8a-f4aee2770406","difficultyBand":"easy","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"2ca2c296-d654-40f2-ad76-ce7f268bd354","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":false},{"id":"0823927a-4951-4226-a49b-adcb690760f5","difficultyBand":"medium","calculatorUse":null,"totalMarks":7,"questionType":"LA","isCalculation":false},{"id":"0f13d61c-9b51-4ddd-8a76-5b68c239d9fd","difficultyBand":"easy","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":false},{"id":"307fa360-020f-40ef-956d-95417e3ceac8","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":false},{"id":"3eacbbf9-f07a-4d1f-8856-766abf8a4f98","difficultyBand":"easy","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"77078f32-6374-42f8-b1d3-415993228c21","difficultyBand":"medium","calculatorUse":null,"totalMarks":7,"questionType":"LA","isCalculation":false},{"id":"82d47be0-9de4-468e-b752-a583b4bc530e","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":false},{"id":"8ad8132c-7132-47ed-984e-330d39b9377b","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":false},{"id":"946e6230-7a3d-4000-b391-7ce2b0b2fff0","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":false},{"id":"95db6ab5-e80c-4d2d-a2cf-1cc236927545","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":false},{"id":"bf86c7a2-b153-466c-8c35-753baa794d98","difficultyBand":"medium","calculatorUse":null,"totalMarks":7,"questionType":"LA","isCalculation":false},{"id":"c213831d-487c-4cf4-8596-a3d9a20e1a71","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":false},{"id":"d929e877-ba83-4854-ad2c-efd7438df34b","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"e2a4f4b3-0e2e-4895-ba94-4348ce074c2c","difficultyBand":"easy","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"1000ed6d-fe65-4492-9d9e-726a40b78c5a","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"4880f865-59f2-4f22-a907-7a0cee1f38ec","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":false},{"id":"157ceeaa-f5d8-4b13-8520-5772c26fad8f","difficultyBand":"medium","calculatorUse":null,"totalMarks":7,"questionType":"LA","isCalculation":false},{"id":"16b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Integration by substitution (A-level only)","topicId":"7d4c8fb4-456a-4cfc-a419-05fa7441e2a1","topicSidebar":"$L3c","topicSlug":"integration-by-substitution","topicTitle":"1.8.8 Integration by substitution (A-level only)"}]}],"$L3d","$L3e"]}]

1.8.8 Integration by substitution (A-level only)

Question 9

1.8.8 Integration by substitution (A-level only) Questions

1.8.8 Integration by substitution (A-level only)

Question 9

1.8.8 Integration by substitution (A-level only) Questions