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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":12,"initialQuestionWindow":[{"id":"7242d354-9652-4ce4-b93c-a86a27c4f362","revisionId":"b957f197-4c86-4b3b-96d2-9f36989e55e0","questionNumber":null,"questionType":"LA","specification":"","difficulty":2,"difficultyBand":"easy","tier":null,"totalMarks":5,"sourceTitle":null,"calculatorUse":null,"isCalculation":false,"parts":[{"id":"779b6cfe-e921-4545-a5d0-674afa30e173","label":"a","content":"Find: $\\int_0^2 \\frac{12}{(3x + 2)^2} dx$","markscheme":"- Rewrites as $12(3x+2)^{-2}$ and integrates: $\\left[ 12(3x+2)^{-1} \\times \\frac{1}{3} \\times (-1) \\right]_0^2 = \\left[ -4(3x+2)^{-1} \\right]_0^2$ M1 A1\n- Substitutes limits: $\\left[ -4(8)^{-1} \\right] - \\left[ -4(2)^{-1} \\right]$ M1\n- Calculates values: $-\\frac{1}{2} + 2$ M1\n- Final answer $\\frac{3}{2}$ or $1.5$ A1","marks":5,"order":0,"markingRubric":{"totalMarks":5,"points":[{"id":"p1","marks":1,"description":"Evidence of rewriting the integrand as a power and attempting to integrate to the form $k(3x+2)^{-1}$.","code":"M"},{"id":"p2","marks":1,"description":"Correct integration including the factor of 1/3, resulting in $-4(3x+2)^{-1}$.","code":"A"},{"id":"p3","marks":1,"description":"Substitution of the limits 2 and 0 into their integrated expression.","code":"M"},{"id":"p4","marks":1,"description":"Correct evaluation of both limit terms before final simplification.","code":"M"},{"id":"p5","marks":1,"description":"Final simplified answer.","code":"A","acceptable":["3/2","1.5"]}],"finalAnswers":["3/2"],"generatedAt":"2026-05-06T14:04:10.493Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"direct_integration","label":"Direct Integration (Reverse Chain Rule)","steps":[{"type":"text","order":0,"title":"Rewrite in index form","content":"Before integrating, it is helpful to rewrite the fraction in index form (with a negative exponent). \n\nWe rewrite the integrand $\\frac{12}{(3x+2)^2}$ as:\n\n$$12(3x+2)^{-2}$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\int 12(\\htmlData{anchor=linear-term}{\\color{@sky}{3x+2}})^{-2} dx$"},{"latex":"$$= \\frac{12}{-1}(\\color{@sky}{3x+2})^{-1} \\times \\htmlData{anchor=recip-deriv}{\\color{@amber}{\\frac{1}{3}}}$"},{"latex":"$$= \\htmlData{anchor=simplified-term}{\\color{@purple}{-4(3x+2)^{-1}}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"linear-term"},{"color":"amber","style":"text-color","anchor":"recip-deriv"},{"color":"purple","style":"text-color","anchor":"simplified-term"}],"connections":[{"to":"recip-deriv","from":"linear-term","color":"amber","label":"divide by x coefficient","style":"dashed"}]},"order":1,"title":"Apply reverse chain rule","description":"To integrate $12(3x+2)^{-2}$ using the reverse chain rule, we raise the power of the bracket by 1 to get $-1$, divide by this new power, and also divide by the derivative of the inside bracket (which is $3$)."},{"type":"text","order":2,"title":"Set up definite integral","content":"Now we express the result using limits from $0$ to $2$. This earns the first method and accuracy marks ($M1$ and $A1$):\n\n$$\\left[ -4(3x+2)^{-1} \\right]_0^2 = \\left[ -\\frac{4}{3x+2} \\right]_0^2$$"},{"type":"text","order":3,"title":"Substitute the limits","content":"Next, substitute the upper limit $x = 2$ and the lower limit $x = 0$ into the expression ($M1$):\n\n$$\\left( -\\frac{4}{3(2)+2} \\right) - \\left( -\\frac{4}{3(0)+2} \\right)$$\n\nSimplify the terms inside the brackets:\n\n$$\\left( -\\frac{4}{8} \\right) - \\left( -\\frac{4}{2} \\right)$$"},{"type":"text","order":4,"title":"Calculate final value","content":"Simplify the fractions and calculate the final result ($M1$):\n\n$$-\\frac{1}{2} - (-2) = -\\frac{1}{2} + 2 = \\frac{3}{2}$$\n\nThus, the final evaluation of the definite integral is $\\frac{3}{2}$ or $1.5$ ($A1$).","calculator":[{"gdc":{"expression":"\\int(12/(3x+2)^2, 0, 2)","instructions":"Go to the Run-Matrix menu, press [Option], then [Calculus] (F4), then [d/dx] (F4) or [Integral] (F4 depending on OS). Enter the integral expression with lower limit 0 and upper limit 2."},"ti84":{"expression":"fnInt(12/(3X+2)^2, X, 0, 2)","instructions":"Press [MATH] and select 9:fnInt(. Enter the function, variable, and limits as fnInt(12/(3X+2)^2, X, 0, 2) and press [ENTER]."},"desmos":{"expression":"\\int_{0}^{2}\\frac{12}{\\left(3x+2\\right)^{2}}dx","instructions":"Type 'int' to generate the integral symbol, input 0 as the lower limit and 2 as the upper limit, then type the integrand: 12/(3x+2)^2 dx."},"description":"Evaluate the definite integral using a graphing calculator."}]}]},{"id":"u_substitution","label":"Integration by u-Substitution","steps":[{"type":"text","order":0,"title":"Define the substitution","content":"To solve this integral using substitution, we choose a new variable, $u$, to simplify the integrand.\n\nLet:\n$$u = 3x + 2$$\n\nNext, we find the derivative of $u$ with respect to $x$ to express $dx$ in terms of $du$:\n$$\\frac{du}{dx} = 3 \\implies dx = \\frac{1}{3} du$$"},{"type":"text","order":1,"title":"Transform the limits","content":"Since we are changing the variable from $x$ to $u$, we must also transform the limits of integration ($x = 0$ and $x = 2$) using our substitution formula $u = 3x + 2$:\n\n* **Lower limit:** When $x = 0$:\n $$u = 3(0) + 2 = 2$$\n* **Upper limit:** When $x = 2$:\n $$u = 3(2) + 2 = 8$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$$\\int_{\\htmlData{anchor=low-x}{0}}^{\\htmlData{anchor=up-x}{2}} \\frac{12}{(\\htmlData{anchor=sub-x}{3x + 2})^2} \\, \\htmlData{anchor=dx-orig}{dx}$$"},{"latex":"$$$\\int_{\\htmlData{anchor=low-u}{2}}^{\\htmlData{anchor=up-u}{8}} \\frac{12}{\\htmlData{anchor=sub-u}{u}^2} \\left( \\htmlData{anchor=dx-new}{\\frac{1}{3} du} \\right)$$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"low-u"},{"color":"pink","style":"text-color","anchor":"up-u"},{"color":"amber","style":"text-color","anchor":"sub-u"},{"color":"purple","style":"text-color","anchor":"dx-new"}],"connections":[{"to":"low-u","from":"low-x","color":"sky","label":"lower limit","style":"solid"},{"to":"up-u","from":"up-x","color":"pink","label":"upper limit","style":"solid"},{"to":"sub-u","from":"sub-x","color":"amber","label":"u = 3x+2","style":"solid"},{"to":"dx-new","from":"dx-orig","color":"purple","label":"dx = 1/3 du","style":"solid"}]},"order":2,"title":"Rewrite the integral","description":"Substitute $u$, $dx$, and the new limits of integration into the original integral expression."},{"type":"text","order":3,"title":"Simplify the integrand","content":"Let's simplify the constant factors in our new integral expression:\n\n$$\\int_{2}^{8} \\frac{12}{u^2} \\cdot \\frac{1}{3} \\, du = \\int_{2}^{8} 4u^{-2} \\, du$$"},{"type":"text","order":4,"title":"Integrate with respect to u","content":"Now we integrate $4u^{-2}$ with respect to $u$ by applying the power rule of integration, $\\int u^n \\, du = \\frac{u^{n+1}}{n+1}$:\n\n$$\\int 4u^{-2} \\, du = \\left[ \\frac{4u^{-1}}{-1} \\right] = \\left[ -4u^{-1} \\right]$$\n\nApplying this to our definite integral limits from $2$ to $8$:\n$$\\left[ -\\frac{4}{u} \\right]_{2}^{8}$$"},{"type":"text","order":5,"title":"Evaluate the limits","content":"Substitute the upper limit $8$ and lower limit $2$ into the integrated expression:\n\n$$\\left( -\\frac{4}{8} \\right) - \\left( -\\frac{4}{2} \\right)$$\n\nSimplify each term:\n$$= -\\frac{1}{2} - (-2)$$\n$$= -\\frac{1}{2} + 2 = \\frac{3}{2}$$\n\nSo, the final value of the definite integral is **$\\frac{3}{2}$** (or **$1.5$**).","calculator":[{"gdc":{"expression":"∫(12/(3x+2)², 0, 2)","instructions":"Go to the Run-Matrix menu, press [OPTN], [F4] (CALC), [F4] (∫dx). Enter the function 12/(3x+2)², set the lower bound as 0, and the upper bound as 2."},"ti84":{"expression":"fnInt(12/(3X+2)², X, 0, 2)","instructions":"Press [MATH], select 9:fnInt(. Input the parameters as fnInt(12/(3X+2)², X, 0, 2) and press [ENTER]."},"desmos":{"expression":"\\int_{0}^{2}\\frac{12}{\\left(3x+2\\right)^{2}}dx","instructions":"Type 'int' to create an integral symbol. Put 0 in the lower limit box and 2 in the upper limit box. Inside the integral, type '12/(3x+2)^2 dx'."},"description":"Verify the value of the definite integral using a graphing calculator."}]}]}],"generatedAt":"2026-06-09T06:04:26.203Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"779b6cfe-e921-4545-a5d0-674afa30e173:direct_integration","label":"Direct Integration (Reverse Chain Rule)","steps":[{"type":"text","order":0,"title":"Rewrite in index form","content":"Before integrating, it is helpful to rewrite the fraction in index form (with a negative exponent). \n\nWe rewrite the integrand $\\frac{12}{(3x+2)^2}$ as:\n\n$$12(3x+2)^{-2}$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\int 12(\\htmlData{anchor=linear-term}{\\color{@sky}{3x+2}})^{-2} dx$"},{"latex":"$$= \\frac{12}{-1}(\\color{@sky}{3x+2})^{-1} \\times \\htmlData{anchor=recip-deriv}{\\color{@amber}{\\frac{1}{3}}}$"},{"latex":"$$= \\htmlData{anchor=simplified-term}{\\color{@purple}{-4(3x+2)^{-1}}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"linear-term"},{"color":"amber","style":"text-color","anchor":"recip-deriv"},{"color":"purple","style":"text-color","anchor":"simplified-term"}],"connections":[{"to":"recip-deriv","from":"linear-term","color":"amber","label":"divide by x coefficient","style":"dashed"}]},"order":1,"title":"Apply reverse chain rule","description":"To integrate $12(3x+2)^{-2}$ using the reverse chain rule, we raise the power of the bracket by 1 to get $-1$, divide by this new power, and also divide by the derivative of the inside bracket (which is $3$)."},{"type":"text","order":2,"title":"Set up definite integral","content":"Now we express the result using limits from $0$ to $2$. This earns the first method and accuracy marks ($M1$ and $A1$):\n\n$$\\left[ -4(3x+2)^{-1} \\right]_0^2 = \\left[ -\\frac{4}{3x+2} \\right]_0^2$$"},{"type":"text","order":3,"title":"Substitute the limits","content":"Next, substitute the upper limit $x = 2$ and the lower limit $x = 0$ into the expression ($M1$):\n\n$$\\left( -\\frac{4}{3(2)+2} \\right) - \\left( -\\frac{4}{3(0)+2} \\right)$$\n\nSimplify the terms inside the brackets:\n\n$$\\left( -\\frac{4}{8} \\right) - \\left( -\\frac{4}{2} \\right)$$"},{"type":"text","order":4,"title":"Calculate final value","content":"Simplify the fractions and calculate the final result ($M1$):\n\n$$-\\frac{1}{2} - (-2) = -\\frac{1}{2} + 2 = \\frac{3}{2}$$\n\nThus, the final evaluation of the definite integral is $\\frac{3}{2}$ or $1.5$ ($A1$).","calculator":[{"gdc":{"expression":"\\int(12/(3x+2)^2, 0, 2)","instructions":"Go to the Run-Matrix menu, press [Option], then [Calculus] (F4), then [d/dx] (F4) or [Integral] (F4 depending on OS). Enter the integral expression with lower limit 0 and upper limit 2."},"ti84":{"expression":"fnInt(12/(3X+2)^2, X, 0, 2)","instructions":"Press [MATH] and select 9:fnInt(. Enter the function, variable, and limits as fnInt(12/(3X+2)^2, X, 0, 2) and press [ENTER]."},"desmos":{"expression":"\\int_{0}^{2}\\frac{12}{\\left(3x+2\\right)^{2}}dx","instructions":"Type 'int' to generate the integral symbol, input 0 as the lower limit and 2 as the upper limit, then type the integrand: 12/(3x+2)^2 dx."},"description":"Evaluate the definite integral using a graphing calculator."}]}]},{"id":"779b6cfe-e921-4545-a5d0-674afa30e173:u_substitution","label":"Integration by u-Substitution","steps":[{"type":"text","order":0,"title":"Define the substitution","content":"To solve this integral using substitution, we choose a new variable, $u$, to simplify the integrand.\n\nLet:\n$$u = 3x + 2$$\n\nNext, we find the derivative of $u$ with respect to $x$ to express $dx$ in terms of $du$:\n$$\\frac{du}{dx} = 3 \\implies dx = \\frac{1}{3} du$$"},{"type":"text","order":1,"title":"Transform the limits","content":"Since we are changing the variable from $x$ to $u$, we must also transform the limits of integration ($x = 0$ and $x = 2$) using our substitution formula $u = 3x + 2$:\n\n* **Lower limit:** When $x = 0$:\n $$u = 3(0) + 2 = 2$$\n* **Upper limit:** When $x = 2$:\n $$u = 3(2) + 2 = 8$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$$\\int_{\\htmlData{anchor=low-x}{0}}^{\\htmlData{anchor=up-x}{2}} \\frac{12}{(\\htmlData{anchor=sub-x}{3x + 2})^2} \\, \\htmlData{anchor=dx-orig}{dx}$$"},{"latex":"$$$\\int_{\\htmlData{anchor=low-u}{2}}^{\\htmlData{anchor=up-u}{8}} \\frac{12}{\\htmlData{anchor=sub-u}{u}^2} \\left( \\htmlData{anchor=dx-new}{\\frac{1}{3} du} \\right)$$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"low-u"},{"color":"pink","style":"text-color","anchor":"up-u"},{"color":"amber","style":"text-color","anchor":"sub-u"},{"color":"purple","style":"text-color","anchor":"dx-new"}],"connections":[{"to":"low-u","from":"low-x","color":"sky","label":"lower limit","style":"solid"},{"to":"up-u","from":"up-x","color":"pink","label":"upper limit","style":"solid"},{"to":"sub-u","from":"sub-x","color":"amber","label":"u = 3x+2","style":"solid"},{"to":"dx-new","from":"dx-orig","color":"purple","label":"dx = 1/3 du","style":"solid"}]},"order":2,"title":"Rewrite the integral","description":"Substitute $u$, $dx$, and the new limits of integration into the original integral expression."},{"type":"text","order":3,"title":"Simplify the integrand","content":"Let's simplify the constant factors in our new integral expression:\n\n$$\\int_{2}^{8} \\frac{12}{u^2} \\cdot \\frac{1}{3} \\, du = \\int_{2}^{8} 4u^{-2} \\, du$$"},{"type":"text","order":4,"title":"Integrate with respect to u","content":"Now we integrate $4u^{-2}$ with respect to $u$ by applying the power rule of integration, $\\int u^n \\, du = \\frac{u^{n+1}}{n+1}$:\n\n$$\\int 4u^{-2} \\, du = \\left[ \\frac{4u^{-1}}{-1} \\right] = \\left[ -4u^{-1} \\right]$$\n\nApplying this to our definite integral limits from $2$ to $8$:\n$$\\left[ -\\frac{4}{u} \\right]_{2}^{8}$$"},{"type":"text","order":5,"title":"Evaluate the limits","content":"Substitute the upper limit $8$ and lower limit $2$ into the integrated expression:\n\n$$\\left( -\\frac{4}{8} \\right) - \\left( -\\frac{4}{2} \\right)$$\n\nSimplify each term:\n$$= -\\frac{1}{2} - (-2)$$\n$$= -\\frac{1}{2} + 2 = \\frac{3}{2}$$\n\nSo, the final value of the definite integral is **$\\frac{3}{2}$** (or **$1.5$**).","calculator":[{"gdc":{"expression":"∫(12/(3x+2)², 0, 2)","instructions":"Go to the Run-Matrix menu, press [OPTN], [F4] (CALC), [F4] (∫dx). Enter the function 12/(3x+2)², set the lower bound as 0, and the upper bound as 2."},"ti84":{"expression":"fnInt(12/(3X+2)², X, 0, 2)","instructions":"Press [MATH], select 9:fnInt(. Input the parameters as fnInt(12/(3X+2)², X, 0, 2) and press [ENTER]."},"desmos":{"expression":"\\int_{0}^{2}\\frac{12}{\\left(3x+2\\right)^{2}}dx","instructions":"Type 'int' to create an integral symbol. Put 0 in the lower limit box and 2 in the upper limit box. Inside the integral, type '12/(3x+2)^2 dx'."},"description":"Verify the value of the definite integral using a graphing calculator."}]}]}],"generatedAt":"2026-06-09T06:04:39.659Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":{"percent":35},"walkthroughVideo":null},{"id":"cf8a0e30-159c-4ad3-aeee-9692e1ae78ae","revisionId":"35779ecb-815b-47e6-bb77-ea9c6fe47f49","questionNumber":null,"questionType":"LA","specification":"","difficulty":1,"difficultyBand":"easy","tier":null,"totalMarks":5,"sourceTitle":null,"calculatorUse":null,"isCalculation":false,"parts":[{"id":"9dd1e702-ba8d-446c-a1a7-74400e7b94e7","label":"a","content":"Use the substitution $u = x^2 + 5$ to find $\\int 2x(x^2 + 5)^4 dx$","markscheme":"$$u = x^2 + 5 \\implies \\frac{du}{dx} = 2x$ M1\n$\\int 2x(u)^4 \\frac{1}{2x} du = \\int u^4 du$ M1\n$= \\frac{1}{5}u^5 + c$ M1\n$= \\frac{1}{5}(x^2 + 5)^5 + c$ A2","marks":5,"order":0,"markingRubric":{"totalMarks":5,"points":[{"id":"p1","marks":1,"description":"Differentiates the substitution $u = x^2 + 5$ to find $\\frac{du}{dx}$","code":"M"},{"id":"p2","marks":1,"description":"Correctly substitutes $u$ and $dx$ into the integral to eliminate $x$","code":"M"},{"id":"p3","marks":1,"description":"Integrates the resulting power of $u$","code":"M"},{"id":"p4","marks":2,"description":"Replaces $u$ with the original expression in $x$ and includes the constant of integration","code":"A"}],"finalAnswers":["\\frac{1}{5}(x^2 + 5)^5 + c"],"generatedAt":"2026-05-06T14:04:27.121Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"differential_substitution","label":"Direct Differential Substitution","steps":[{"type":"text","order":0,"title":"Identify substitution and differential","content":"The question asks us to use the substitution $u = x^2 + 5$.\n\nFirst, let's find the derivative of $u$ with respect to $x$:\n\n$$\\frac{du}{dx} = 2x$$\n\nIn differential form, this can be written directly as:\n\n$$du = 2x\\,dx$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\int \\htmlData{anchor=diff-comp}{\\color{@sky}{2x}} \\left(\\htmlData{anchor=u-inner}{\\color{@amber}{x^2 + 5}}\\right)^4 \\htmlData{anchor=dx-comp}{\\color{@sky}{dx}}$"},{"latex":"$$\\int \\left(\\htmlData{anchor=u-sub}{\\color{@amber}{u}}\\right)^4 \\htmlData{anchor=du-sub}{\\color{@sky}{du}}$"}],"highlights":[{"color":"amber","style":"box","anchor":"u-inner"},{"color":"sky","style":"box","anchor":"diff-comp"}],"connections":[{"to":"u-sub","from":"u-inner","color":"amber","label":"substitute u","style":"solid"},{"to":"du-sub","from":"diff-comp","color":"sky","label":"substitute du","style":"solid"}]},"order":1,"title":"Substitute into the integral","description":"We rewrite the original integral to group the differential terms together:\n\n$$\\int (x^2 + 5)^4 \\cdot (2x\\,dx)$$\n\nNow, we substitute $u = x^2 + 5$ and the differential component $du = 2x\\,dx$ directly into the integral."},{"type":"text","order":2,"title":"Integrate with respect to u","content":"Now we integrate $u^4$ with respect to $u$ using the standard power rule for integration, $\\int u^n du = \\frac{u^{n+1}}{n+1} + c$:\n\n$$\\int u^4 du = \\frac{1}{5}u^5 + c$$\n\nwhere $c$ is the constant of integration."},{"type":"text","order":3,"title":"Substitute back to x","content":"Finally, replace $u$ with our original expression in terms of $x$, which is $u = x^2 + 5$:\n\n$$\\frac{1}{5}(x^2 + 5)^5 + c$$"}]},{"id":"dx_substitution","label":"Explicit dx Substitution","steps":[{"type":"text","order":0,"title":"Differentiate the substitution","content":"We are given the substitution $u = x^2 + 5$. First, differentiate $u$ with respect to $x$ to find $\\frac{du}{dx}$:\n\n$$\\frac{du}{dx} = 2x$$\n\nThis step helps us connect the differentials $dx$ and $du$, earning you the first method mark."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\frac{\\htmlData{anchor=du-term}{du}}{\\htmlData{anchor=dx-term}{\\color{@sky}{dx}}} = \\htmlData{anchor=deriv-term}{\\color{@amber}{2x}}$"},{"latex":"$$\\htmlData{anchor=dx-rearranged}{\\color{@sky}{dx}} = \\frac{1}{\\htmlData{anchor=deriv-rearranged}{\\color{@amber}{2x}}}\\,\\htmlData{anchor=du-rearranged}{du}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"dx-term"},{"color":"amber","style":"text-color","anchor":"deriv-term"}],"connections":[{"to":"dx-rearranged","from":"dx-term","color":"sky","label":"rearrange","style":"solid"},{"to":"deriv-rearranged","from":"deriv-term","color":"amber","label":"divide","style":"solid"}]},"order":1,"title":"Rearrange to express dx","description":"Now, rearrange the derivative equation to explicitly express $dx$ in terms of $du$ and $x$."},{"type":"text","order":2,"title":"Substitute into the integral","content":"Now we substitute both $u = x^2 + 5$ and $dx = \\frac{1}{2x}\\,du$ into our original integral:\n\n$$\\int 2x(x^2 + 5)^4 \\, dx = \\int 2x \\cdot (u)^4 \\cdot \\left(\\frac{1}{2x}\\,du\\right)$$\n\nNotice that the $2x$ term in the numerator cancels perfectly with the $2x$ in the denominator:\n\n$$\\int \\cancel{2x} \\cdot u^4 \\cdot \\frac{1}{\\cancel{2x}}\\,du = \\int u^4 \\, du$$\n\nThis algebraic simplification earns the second method mark."},{"type":"text","order":3,"title":"Integrate with respect to u","content":"Next, integrate $u^4$ with respect to $u$ using the power rule for integration, $\\int u^n \\, du = \\frac{u^{n+1}}{n+1} + c$:\n\n$$\\int u^4 \\, du = \\frac{1}{5}u^5 + c$$\n\n*Remember to always include the constant of integration, $+ c$.*"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\htmlData{anchor=u-var}{\\color{@purple}{u}} = \\htmlData{anchor=x-expr}{\\color{@purple}{x^2 + 5}}$"},{"latex":"$$\\frac{1}{5}(\\htmlData{anchor=u-sub}{\\color{@purple}{u}})^5 + c = \\frac{1}{5}(\\htmlData{anchor=x-sub}{\\color{@purple}{x^2 + 5}})^5 + c$"}],"highlights":[{"color":"purple","style":"text-color","anchor":"u-var"},{"color":"purple","style":"text-color","anchor":"x-expr"}],"connections":[{"to":"x-sub","from":"x-expr","color":"purple","label":"substitute back","style":"solid"}]},"order":4,"title":"Substitute back to x","description":"Finally, substitute the original expression for $u$ back into our integrated result to get the final answer in terms of $x$."}]}],"generatedAt":"2026-06-09T02:00:26.438Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"9dd1e702-ba8d-446c-a1a7-74400e7b94e7:differential_substitution","label":"Direct Differential Substitution","steps":[{"type":"text","order":0,"title":"Identify substitution and differential","content":"The question asks us to use the substitution $u = x^2 + 5$.\n\nFirst, let's find the derivative of $u$ with respect to $x$:\n\n$$\\frac{du}{dx} = 2x$$\n\nIn differential form, this can be written directly as:\n\n$$du = 2x\\,dx$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\int \\htmlData{anchor=diff-comp}{\\color{@sky}{2x}} \\left(\\htmlData{anchor=u-inner}{\\color{@amber}{x^2 + 5}}\\right)^4 \\htmlData{anchor=dx-comp}{\\color{@sky}{dx}}$"},{"latex":"$$\\int \\left(\\htmlData{anchor=u-sub}{\\color{@amber}{u}}\\right)^4 \\htmlData{anchor=du-sub}{\\color{@sky}{du}}$"}],"highlights":[{"color":"amber","style":"box","anchor":"u-inner"},{"color":"sky","style":"box","anchor":"diff-comp"}],"connections":[{"to":"u-sub","from":"u-inner","color":"amber","label":"substitute u","style":"solid"},{"to":"du-sub","from":"diff-comp","color":"sky","label":"substitute du","style":"solid"}]},"order":1,"title":"Substitute into the integral","description":"We rewrite the original integral to group the differential terms together:\n\n$$\\int (x^2 + 5)^4 \\cdot (2x\\,dx)$$\n\nNow, we substitute $u = x^2 + 5$ and the differential component $du = 2x\\,dx$ directly into the integral."},{"type":"text","order":2,"title":"Integrate with respect to u","content":"Now we integrate $u^4$ with respect to $u$ using the standard power rule for integration, $\\int u^n du = \\frac{u^{n+1}}{n+1} + c$:\n\n$$\\int u^4 du = \\frac{1}{5}u^5 + c$$\n\nwhere $c$ is the constant of integration."},{"type":"text","order":3,"title":"Substitute back to x","content":"Finally, replace $u$ with our original expression in terms of $x$, which is $u = x^2 + 5$:\n\n$$\\frac{1}{5}(x^2 + 5)^5 + c$$"}]},{"id":"9dd1e702-ba8d-446c-a1a7-74400e7b94e7:dx_substitution","label":"Explicit dx Substitution","steps":[{"type":"text","order":0,"title":"Differentiate the substitution","content":"We are given the substitution $u = x^2 + 5$. First, differentiate $u$ with respect to $x$ to find $\\frac{du}{dx}$:\n\n$$\\frac{du}{dx} = 2x$$\n\nThis step helps us connect the differentials $dx$ and $du$, earning you the first method mark."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\frac{\\htmlData{anchor=du-term}{du}}{\\htmlData{anchor=dx-term}{\\color{@sky}{dx}}} = \\htmlData{anchor=deriv-term}{\\color{@amber}{2x}}$"},{"latex":"$$\\htmlData{anchor=dx-rearranged}{\\color{@sky}{dx}} = \\frac{1}{\\htmlData{anchor=deriv-rearranged}{\\color{@amber}{2x}}}\\,\\htmlData{anchor=du-rearranged}{du}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"dx-term"},{"color":"amber","style":"text-color","anchor":"deriv-term"}],"connections":[{"to":"dx-rearranged","from":"dx-term","color":"sky","label":"rearrange","style":"solid"},{"to":"deriv-rearranged","from":"deriv-term","color":"amber","label":"divide","style":"solid"}]},"order":1,"title":"Rearrange to express dx","description":"Now, rearrange the derivative equation to explicitly express $dx$ in terms of $du$ and $x$."},{"type":"text","order":2,"title":"Substitute into the integral","content":"Now we substitute both $u = x^2 + 5$ and $dx = \\frac{1}{2x}\\,du$ into our original integral:\n\n$$\\int 2x(x^2 + 5)^4 \\, dx = \\int 2x \\cdot (u)^4 \\cdot \\left(\\frac{1}{2x}\\,du\\right)$$\n\nNotice that the $2x$ term in the numerator cancels perfectly with the $2x$ in the denominator:\n\n$$\\int \\cancel{2x} \\cdot u^4 \\cdot \\frac{1}{\\cancel{2x}}\\,du = \\int u^4 \\, du$$\n\nThis algebraic simplification earns the second method mark."},{"type":"text","order":3,"title":"Integrate with respect to u","content":"Next, integrate $u^4$ with respect to $u$ using the power rule for integration, $\\int u^n \\, du = \\frac{u^{n+1}}{n+1} + c$:\n\n$$\\int u^4 \\, du = \\frac{1}{5}u^5 + c$$\n\n*Remember to always include the constant of integration, $+ c$.*"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\htmlData{anchor=u-var}{\\color{@purple}{u}} = \\htmlData{anchor=x-expr}{\\color{@purple}{x^2 + 5}}$"},{"latex":"$$\\frac{1}{5}(\\htmlData{anchor=u-sub}{\\color{@purple}{u}})^5 + c = \\frac{1}{5}(\\htmlData{anchor=x-sub}{\\color{@purple}{x^2 + 5}})^5 + c$"}],"highlights":[{"color":"purple","style":"text-color","anchor":"u-var"},{"color":"purple","style":"text-color","anchor":"x-expr"}],"connections":[{"to":"x-sub","from":"x-expr","color":"purple","label":"substitute back","style":"solid"}]},"order":4,"title":"Substitute back to x","description":"Finally, substitute the original expression for $u$ back into our integrated result to get the final answer in terms of $x$."}]}],"generatedAt":"2026-06-09T02:00:49.640Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null},{"id":"1c062306-1388-4be0-af8a-f4aee2770406","revisionId":"33729c02-7b10-4ee5-b7ae-6a9961ed3d28","questionNumber":"2","questionType":"LA","specification":"Integrate with respect to $x$:","difficulty":1,"difficultyBand":"easy","tier":null,"totalMarks":6,"sourceTitle":"Integration","calculatorUse":null,"isCalculation":true,"parts":[{"id":"0d1dbc9c-618f-4220-9924-d83058c1f13e","label":"a","content":"$$\\sqrt{5x - 3}$","markscheme":"- Rewrites as $(5x-3)^{\\frac{1}{2}}$ and integrates: $\\frac{2}{3}(5x-3)^{\\frac{3}{2}} \\times \\frac{1}{5} + C$ M1\n- Simplifies to $\\frac{2}{15}(5x-3)^{\\frac{3}{2}} + C$ A1\n- Includes constant of integration $+ C$ B1","marks":3,"order":0,"markingRubric":{"totalMarks":3,"points":[{"id":"p1","marks":1,"description":"Integrates to form $k(5x-3)^{\\frac{3}{2}}$.","code":"M"},{"id":"p2","marks":1,"description":"Correctly applies the factor of $\\frac{1}{5}$.","code":"M"},{"id":"p3","marks":1,"description":"Final answer $\\frac{2}{15}(5x-3)^{\\frac{3}{2}} + C$.","code":"A"}],"finalAnswers":["\\frac{2}{15}(5x-3)^{\\frac{3}{2}} + C"],"generatedAt":"2026-05-06T03:37:59.137Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"reverse-chain-rule","label":"Reverse Chain Rule","steps":[{"type":"text","order":0,"title":"Rewrite the integral","content":"To integrate $\\sqrt{5x - 3}$ with respect to $x$, we first write it in index form. Recall that a square root can be written as a fractional exponent of $\\frac{1}{2}$:\n\n$$\\int \\sqrt{5x - 3} \\, dx = \\int (5x - 3)^{\\frac{1}{2}} \\, dx$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$$\\int (\\htmlData{anchor=linear}{\\color{@sky}{5x - 3}})^{\\htmlData{anchor=power-old}{\\color{@amber}{\\frac{1}{2}}}} \\, dx$$"},{"latex":"$$$= \\frac{(\\color{@sky}{5x - 3})^{\\htmlData{anchor=power-new}{\\color{@amber}{\\frac{3}{2}}}}}{\\color{@amber}{\\frac{3}{2}}} \\times \\frac{1}{\\htmlData{anchor=coeff}{\\color{@purple}{5}}} + C$$"}],"highlights":[{"color":"amber","style":"text-color","anchor":"power-old"},{"color":"amber","style":"text-color","anchor":"power-new"},{"color":"purple","style":"text-color","anchor":"coeff"}],"connections":[{"to":"power-new","from":"power-old","color":"amber","label":"Add 1 to power","style":"solid"},{"to":"coeff","from":"linear","color":"purple","label":"Divide by coefficient","style":"solid"}]},"order":1,"title":"Apply reverse chain rule","description":"We apply the reverse chain rule for linear compositions $\\int (ax + b)^n \\, dx = \\frac{(ax + b)^{n+1}}{a(n+1)} + C$. Here, we increase the power by $1$ to get $\\frac{3}{2}$, divide by this new power, and also divide by the coefficient of $x$, which is $5$."},{"type":"text","order":2,"title":"Simplify and add constant","content":"Let's simplify the constant coefficient of our integrated expression. \n\nDividing by $\\frac{3}{2}$ is equivalent to multiplying by $\\frac{2}{3}$. Let's combine this with the division by $5$:\n\n$$\\frac{2}{3} \\times \\frac{1}{5} = \\frac{2}{15}$$\n\nPutting it all together, we get our final integrated expression, remembering to include the constant of integration, $+ C$:\n\n$$\\frac{2}{15}(5x - 3)^{\\frac{3}{2}} + C$$"},{"type":"text","order":3,"title":"Markscheme breakdown","content":"Here is how the **3 marks** are awarded for this question:\n\n* **M1**: For rewriting the integrand as $(5x-3)^{\\frac{1}{2}}$ and attempting to integrate to get $\\frac{2}{3}(5x-3)^{\\frac{3}{2}} \\times \\frac{1}{5}$.\n* **A1**: For correctly simplifying the expression to $\\frac{2}{15}(5x-3)^{\\frac{3}{2}}$.\n* **B1**: For including the constant of integration, $+ C$."}]},{"id":"substitution","label":"Substitution Method","steps":[{"type":"text","order":0,"title":"Set up the integral","content":"We want to find the indefinite integral of $\\sqrt{5x - 3}$ with respect to $x$. First, let's write it down using the integration symbol and express the square root as a fractional power:\n\n$$\\int \\sqrt{5x - 3} \\, dx = \\int (5x - 3)^{\\frac{1}{2}} \\, dx$$","highlights":[{"text":"$$\\sqrt{5x - 3}$","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"u = \\htmlData{anchor=sub-expr}{\\color{@sky}{5x - 3}}"},{"text":"Differentiating both sides with respect to $x$ gives:"},{"latex":"\\frac{du}{dx} = 5"},{"text":"Which we can rewrite to find an expression for $dx$:"},{"latex":"dx = \\htmlData{anchor=dx-expr}{\\color{@amber}{\\frac{1}{5} \\, du}}"}],"highlights":[{"color":"sky","style":"box","anchor":"sub-expr"},{"color":"amber","style":"box","anchor":"dx-expr"}],"connections":[]},"order":1,"title":"Choose the substitution","description":"We choose $u = 5x - 3$ to simplify the expression inside the square root."},{"type":"hyper_latex","block":{"nodes":[{"latex":"\\int (\\htmlData{anchor=u-term}{\\color{@sky}{5x - 3}})^{\\frac{1}{2}} \\, \\htmlData{anchor=dx-term}{\\color{@amber}{dx}}"},{"text":"Substituting our terms for $u$ and $dx$ yields:"},{"latex":"\\int \\htmlData{anchor=u-sub}{\\color{@sky}{u}}^{\\frac{1}{2}} \\cdot \\htmlData{anchor=dx-sub}{\\color{@amber}{\\frac{1}{5} \\, du}} = \\frac{1}{5} \\int u^{\\frac{1}{2}} \\, du"}],"highlights":[{"color":"sky","style":"text-color","anchor":"u-term"},{"color":"amber","style":"text-color","anchor":"dx-term"}],"connections":[{"to":"u-sub","from":"u-term","color":"sky","label":"substitute u","style":"solid"},{"to":"dx-sub","from":"dx-term","color":"amber","label":"substitute dx","style":"solid"}]},"order":2,"title":"Transform the integral","description":"Now, substitute $u$ and $dx$ into the original integral to write it entirely in terms of $u$."},{"type":"text","order":3,"title":"Integrate with respect to u","content":"Now we integrate the power of $u$ using the standard power rule for integration, $\\int u^n \\, du = \\frac{u^{n+1}}{n+1} + C$:\n\n$$\\frac{1}{5} \\int u^{\\frac{1}{2}} \\, du = \\frac{1}{5} \\left( \\frac{u^{\\frac{3}{2}}}{\\frac{3}{2}} \\right) + C$$\n\nSimplifying the fraction:\n\n$$\\frac{1}{5} \\left( \\frac{2}{3} u^{\\frac{3}{2}} \\right) + C = \\frac{2}{15} u^{\\frac{3}{2}} + C$$\n\n*(Note: Remembering to include the constant of integration $+ C$ is required to earn full marks.)*"},{"type":"hyper_latex","block":{"nodes":[{"latex":"\\frac{2}{15} \\htmlData{anchor=u-back}{\\color{@sky}{u}}^{\\frac{3}{2}} + C"},{"text":"Replacing $u$ with our original expression $5x - 3$ gives:"},{"latex":"\\frac{2}{15} (\\htmlData{anchor=x-back}{\\color{@sky}{5x - 3}})^{\\frac{3}{2}} + C"}],"highlights":[{"color":"sky","style":"text-color","anchor":"u-back"}],"connections":[{"to":"x-back","from":"u-back","color":"sky","label":"substitute back","style":"solid"}]},"order":4,"title":"Substitute back to x","description":"Substitute $u = 5x - 3$ back into the integrated expression to get the final answer in terms of $x$."}]}],"generatedAt":"2026-06-09T03:32:19.011Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]},{"id":"89f7c1cd-8cab-4a4b-852d-5f1bb4ab1080","label":"b","content":"$$e^{2x+3}$","markscheme":"- Integrates exponential: $\\frac{1}{2}e^{2x+3} + C$ M1\n- Correct coefficient $\\frac{1}{2}$ A1\n- Includes constant of integration $+ C$ B1","marks":3,"order":1,"markingRubric":{"totalMarks":3,"points":[{"id":"p1","marks":1,"description":"Recognises integral of $e^{ax+b}$ is $\\frac{1}{a}e^{ax+b}$.","code":"M"},{"id":"p2","marks":1,"description":"Correctly identifies $a=2$.","code":"A"},{"id":"p3","marks":1,"description":"Final answer $\\frac{1}{2}e^{2x+3} + C$.","code":"B"}],"finalAnswers":["\\frac{1}{2}e^{2x+3} + C"],"generatedAt":"2026-05-06T03:37:59.137Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"reverse-chain-rule","label":"Reverse Chain Rule","steps":[{"type":"text","order":0,"title":"Set up the integral","content":"We need to find the indefinite integral of the exponential function:\n\n$$\\int e^{2x+3} \\, dx$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\int e^{\\htmlData{anchor=linear}{\\color{@sky}{2}}x + 3} \\, dx$"},{"latex":"$$= \\frac{1}{\\htmlData{anchor=deriv}{\\color{@sky}{2}}} e^{\\htmlData{anchor=power}{\\color{@sky}{2}x + 3}} + \\htmlData{anchor=c-val}{\\color{@purple}{C}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"linear"},{"color":"sky","style":"text-color","anchor":"deriv"},{"color":"purple","style":"text-color","anchor":"c-val"}],"connections":[{"to":"deriv","from":"linear","color":"sky","label":"divide by derivative","style":"solid"},{"to":"power","from":"linear","color":"sky","label":"power remains","style":"dashed"}]},"order":1,"title":"Apply reverse chain rule","description":"For any exponential function with a linear exponent of the form $ax+b$, we divide the function by the derivative of the exponent, which is the constant $a$."},{"type":"text","order":2,"title":"Summarize marks","content":"Our final answer is:\n\n$$\\frac{1}{2}e^{2x+3} + C$$\n\n### Marks Breakdown\n- **M1**: For integrating the exponential term to get a form of $k e^{2x+3}$.\n- **A1**: For obtaining the correct coefficient of $\\frac{1}{2}$.\n- **B1**: For including the constant of integration, $+ C$."}]},{"id":"substitution","label":"Integration by Substitution","steps":[{"type":"text","order":0,"title":"Set up the integral","content":"We need to integrate the function $e^{2x+3}$ with respect to $x$. Let's write this as an indefinite integral:\n\n$$\\int e^{2x+3} \\, dx$$"},{"type":"hyper_latex","block":{"nodes":[{"text":"Let's choose our substitution variable $u$ to represent the linear expression in the exponent:"},{"latex":"$$\\htmlData{anchor=u-def}{\\color{@sky}{u}} = \\htmlData{anchor=exponent}{\\color{@sky}{2x + 3}}$"},{"text":"Now, we differentiate $u$ with respect to $x$ to find the differential term:"},{"latex":"$$\\frac{d\\htmlData{anchor=u-diff}{\\color{@sky}{u}}}{dx} = \\htmlData{anchor=deriv}{\\color{@amber}{2}}$"},{"text":"Rearranging this expression gives us the relationship for $dx$:"},{"latex":"$$\\htmlData{anchor=dx-val}{\\color{@purple}{dx}} = \\frac{1}{\\htmlData{anchor=deriv-sub}{\\color{@amber}{2}}} \\, \\htmlData{anchor=du-val}{\\color{@purple}{du}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"u-def"},{"color":"sky","style":"text-color","anchor":"exponent"},{"color":"amber","style":"text-color","anchor":"deriv"},{"color":"purple","style":"text-color","anchor":"dx-val"},{"color":"purple","style":"text-color","anchor":"du-val"}],"connections":[{"to":"deriv-sub","from":"deriv","color":"amber","label":"rearrange","style":"solid"}]},"order":1,"title":"Define the substitution","description":"We define a new variable $u$ to simplify the exponent of the exponential function."},{"type":"text","order":2,"title":"Substitute into the integral","content":"Now we substitute both $u = 2x + 3$ and $dx = \\frac{1}{2} \\, du$ back into the original integral:\n\n$$\\int e^{2x+3} \\, dx = \\int e^{u} \\left( \\frac{1}{2} \\, du \\right)$$\n\nBy pulling the constant factor $\\frac{1}{2}$ out of the integration sign, we get:\n\n$$\\frac{1}{2} \\int e^{u} \\, du$$"},{"type":"text","order":3,"title":"Integrate with respect to u","content":"The integral of $e^u$ is a standard integral from our formula booklet:\n\n$$\\int e^u \\, du = e^u + C$$\n\nApplying this standard form, we get:\n\n$$\\frac{1}{2} e^u + C$$"},{"type":"text","order":4,"title":"Substitute back to x","content":"Finally, we substitute $u = 2x+3$ back into our result to express our final answer in terms of $x$:\n\n$$\\frac{1}{2} e^{2x+3} + C$$\n\n### Markscheme Breakdown\n- **M1**: Integrating the exponential function to get the correct general form.\n- **A1**: Finding the correct coefficient $\\frac{1}{2}$.\n- **B1**: Including the constant of integration $+ C$."}]}],"generatedAt":"2026-06-09T03:33:44.353Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"0d1dbc9c-618f-4220-9924-d83058c1f13e:reverse-chain-rule","label":"(a) Reverse Chain Rule","steps":[{"type":"text","order":0,"title":"Rewrite the integral","content":"To integrate $\\sqrt{5x - 3}$ with respect to $x$, we first write it in index form. Recall that a square root can be written as a fractional exponent of $\\frac{1}{2}$:\n\n$$\\int \\sqrt{5x - 3} \\, dx = \\int (5x - 3)^{\\frac{1}{2}} \\, dx$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$$\\int (\\htmlData{anchor=linear}{\\color{@sky}{5x - 3}})^{\\htmlData{anchor=power-old}{\\color{@amber}{\\frac{1}{2}}}} \\, dx$$"},{"latex":"$$$= \\frac{(\\color{@sky}{5x - 3})^{\\htmlData{anchor=power-new}{\\color{@amber}{\\frac{3}{2}}}}}{\\color{@amber}{\\frac{3}{2}}} \\times \\frac{1}{\\htmlData{anchor=coeff}{\\color{@purple}{5}}} + C$$"}],"highlights":[{"color":"amber","style":"text-color","anchor":"power-old"},{"color":"amber","style":"text-color","anchor":"power-new"},{"color":"purple","style":"text-color","anchor":"coeff"}],"connections":[{"to":"power-new","from":"power-old","color":"amber","label":"Add 1 to power","style":"solid"},{"to":"coeff","from":"linear","color":"purple","label":"Divide by coefficient","style":"solid"}]},"order":1,"title":"Apply reverse chain rule","description":"We apply the reverse chain rule for linear compositions $\\int (ax + b)^n \\, dx = \\frac{(ax + b)^{n+1}}{a(n+1)} + C$. Here, we increase the power by $1$ to get $\\frac{3}{2}$, divide by this new power, and also divide by the coefficient of $x$, which is $5$."},{"type":"text","order":2,"title":"Simplify and add constant","content":"Let's simplify the constant coefficient of our integrated expression. \n\nDividing by $\\frac{3}{2}$ is equivalent to multiplying by $\\frac{2}{3}$. Let's combine this with the division by $5$:\n\n$$\\frac{2}{3} \\times \\frac{1}{5} = \\frac{2}{15}$$\n\nPutting it all together, we get our final integrated expression, remembering to include the constant of integration, $+ C$:\n\n$$\\frac{2}{15}(5x - 3)^{\\frac{3}{2}} + C$$"},{"type":"text","order":3,"title":"Markscheme breakdown","content":"Here is how the **3 marks** are awarded for this question:\n\n* **M1**: For rewriting the integrand as $(5x-3)^{\\frac{1}{2}}$ and attempting to integrate to get $\\frac{2}{3}(5x-3)^{\\frac{3}{2}} \\times \\frac{1}{5}$.\n* **A1**: For correctly simplifying the expression to $\\frac{2}{15}(5x-3)^{\\frac{3}{2}}$.\n* **B1**: For including the constant of integration, $+ C$."}]},{"id":"0d1dbc9c-618f-4220-9924-d83058c1f13e:substitution","label":"(a) Substitution Method","steps":[{"type":"text","order":0,"title":"Set up the integral","content":"We want to find the indefinite integral of $\\sqrt{5x - 3}$ with respect to $x$. First, let's write it down using the integration symbol and express the square root as a fractional power:\n\n$$\\int \\sqrt{5x - 3} \\, dx = \\int (5x - 3)^{\\frac{1}{2}} \\, dx$$","highlights":[{"text":"$$\\sqrt{5x - 3}$","location":"specification"}]},{"type":"hyper_latex","block":{"nodes":[{"latex":"u = \\htmlData{anchor=sub-expr}{\\color{@sky}{5x - 3}}"},{"text":"Differentiating both sides with respect to $x$ gives:"},{"latex":"\\frac{du}{dx} = 5"},{"text":"Which we can rewrite to find an expression for $dx$:"},{"latex":"dx = \\htmlData{anchor=dx-expr}{\\color{@amber}{\\frac{1}{5} \\, du}}"}],"highlights":[{"color":"sky","style":"box","anchor":"sub-expr"},{"color":"amber","style":"box","anchor":"dx-expr"}],"connections":[]},"order":1,"title":"Choose the substitution","description":"We choose $u = 5x - 3$ to simplify the expression inside the square root."},{"type":"hyper_latex","block":{"nodes":[{"latex":"\\int (\\htmlData{anchor=u-term}{\\color{@sky}{5x - 3}})^{\\frac{1}{2}} \\, \\htmlData{anchor=dx-term}{\\color{@amber}{dx}}"},{"text":"Substituting our terms for $u$ and $dx$ yields:"},{"latex":"\\int \\htmlData{anchor=u-sub}{\\color{@sky}{u}}^{\\frac{1}{2}} \\cdot \\htmlData{anchor=dx-sub}{\\color{@amber}{\\frac{1}{5} \\, du}} = \\frac{1}{5} \\int u^{\\frac{1}{2}} \\, du"}],"highlights":[{"color":"sky","style":"text-color","anchor":"u-term"},{"color":"amber","style":"text-color","anchor":"dx-term"}],"connections":[{"to":"u-sub","from":"u-term","color":"sky","label":"substitute u","style":"solid"},{"to":"dx-sub","from":"dx-term","color":"amber","label":"substitute dx","style":"solid"}]},"order":2,"title":"Transform the integral","description":"Now, substitute $u$ and $dx$ into the original integral to write it entirely in terms of $u$."},{"type":"text","order":3,"title":"Integrate with respect to u","content":"Now we integrate the power of $u$ using the standard power rule for integration, $\\int u^n \\, du = \\frac{u^{n+1}}{n+1} + C$:\n\n$$\\frac{1}{5} \\int u^{\\frac{1}{2}} \\, du = \\frac{1}{5} \\left( \\frac{u^{\\frac{3}{2}}}{\\frac{3}{2}} \\right) + C$$\n\nSimplifying the fraction:\n\n$$\\frac{1}{5} \\left( \\frac{2}{3} u^{\\frac{3}{2}} \\right) + C = \\frac{2}{15} u^{\\frac{3}{2}} + C$$\n\n*(Note: Remembering to include the constant of integration $+ C$ is required to earn full marks.)*"},{"type":"hyper_latex","block":{"nodes":[{"latex":"\\frac{2}{15} \\htmlData{anchor=u-back}{\\color{@sky}{u}}^{\\frac{3}{2}} + C"},{"text":"Replacing $u$ with our original expression $5x - 3$ gives:"},{"latex":"\\frac{2}{15} (\\htmlData{anchor=x-back}{\\color{@sky}{5x - 3}})^{\\frac{3}{2}} + C"}],"highlights":[{"color":"sky","style":"text-color","anchor":"u-back"}],"connections":[{"to":"x-back","from":"u-back","color":"sky","label":"substitute back","style":"solid"}]},"order":4,"title":"Substitute back to x","description":"Substitute $u = 5x - 3$ back into the integrated expression to get the final answer in terms of $x$."}]},{"id":"89f7c1cd-8cab-4a4b-852d-5f1bb4ab1080:reverse-chain-rule","label":"(b) Reverse Chain Rule","steps":[{"type":"text","order":0,"title":"Set up the integral","content":"We need to find the indefinite integral of the exponential function:\n\n$$\\int e^{2x+3} \\, dx$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\int e^{\\htmlData{anchor=linear}{\\color{@sky}{2}}x + 3} \\, dx$"},{"latex":"$$= \\frac{1}{\\htmlData{anchor=deriv}{\\color{@sky}{2}}} e^{\\htmlData{anchor=power}{\\color{@sky}{2}x + 3}} + \\htmlData{anchor=c-val}{\\color{@purple}{C}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"linear"},{"color":"sky","style":"text-color","anchor":"deriv"},{"color":"purple","style":"text-color","anchor":"c-val"}],"connections":[{"to":"deriv","from":"linear","color":"sky","label":"divide by derivative","style":"solid"},{"to":"power","from":"linear","color":"sky","label":"power remains","style":"dashed"}]},"order":1,"title":"Apply reverse chain rule","description":"For any exponential function with a linear exponent of the form $ax+b$, we divide the function by the derivative of the exponent, which is the constant $a$."},{"type":"text","order":2,"title":"Summarize marks","content":"Our final answer is:\n\n$$\\frac{1}{2}e^{2x+3} + C$$\n\n### Marks Breakdown\n- **M1**: For integrating the exponential term to get a form of $k e^{2x+3}$.\n- **A1**: For obtaining the correct coefficient of $\\frac{1}{2}$.\n- **B1**: For including the constant of integration, $+ C$."}]},{"id":"89f7c1cd-8cab-4a4b-852d-5f1bb4ab1080:substitution","label":"(b) Integration by Substitution","steps":[{"type":"text","order":0,"title":"Set up the integral","content":"We need to integrate the function $e^{2x+3}$ with respect to $x$. Let's write this as an indefinite integral:\n\n$$\\int e^{2x+3} \\, dx$$"},{"type":"hyper_latex","block":{"nodes":[{"text":"Let's choose our substitution variable $u$ to represent the linear expression in the exponent:"},{"latex":"$$\\htmlData{anchor=u-def}{\\color{@sky}{u}} = \\htmlData{anchor=exponent}{\\color{@sky}{2x + 3}}$"},{"text":"Now, we differentiate $u$ with respect to $x$ to find the differential term:"},{"latex":"$$\\frac{d\\htmlData{anchor=u-diff}{\\color{@sky}{u}}}{dx} = \\htmlData{anchor=deriv}{\\color{@amber}{2}}$"},{"text":"Rearranging this expression gives us the relationship for $dx$:"},{"latex":"$$\\htmlData{anchor=dx-val}{\\color{@purple}{dx}} = \\frac{1}{\\htmlData{anchor=deriv-sub}{\\color{@amber}{2}}} \\, \\htmlData{anchor=du-val}{\\color{@purple}{du}}$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"u-def"},{"color":"sky","style":"text-color","anchor":"exponent"},{"color":"amber","style":"text-color","anchor":"deriv"},{"color":"purple","style":"text-color","anchor":"dx-val"},{"color":"purple","style":"text-color","anchor":"du-val"}],"connections":[{"to":"deriv-sub","from":"deriv","color":"amber","label":"rearrange","style":"solid"}]},"order":1,"title":"Define the substitution","description":"We define a new variable $u$ to simplify the exponent of the exponential function."},{"type":"text","order":2,"title":"Substitute into the integral","content":"Now we substitute both $u = 2x + 3$ and $dx = \\frac{1}{2} \\, du$ back into the original integral:\n\n$$\\int e^{2x+3} \\, dx = \\int e^{u} \\left( \\frac{1}{2} \\, du \\right)$$\n\nBy pulling the constant factor $\\frac{1}{2}$ out of the integration sign, we get:\n\n$$\\frac{1}{2} \\int e^{u} \\, du$$"},{"type":"text","order":3,"title":"Integrate with respect to u","content":"The integral of $e^u$ is a standard integral from our formula booklet:\n\n$$\\int e^u \\, du = e^u + C$$\n\nApplying this standard form, we get:\n\n$$\\frac{1}{2} e^u + C$$"},{"type":"text","order":4,"title":"Substitute back to x","content":"Finally, we substitute $u = 2x+3$ back into our result to express our final answer in terms of $x$:\n\n$$\\frac{1}{2} e^{2x+3} + C$$\n\n### Markscheme Breakdown\n- **M1**: Integrating the exponential function to get the correct general form.\n- **A1**: Finding the correct coefficient $\\frac{1}{2}$.\n- **B1**: Including the constant of integration $+ C$."}]}],"generatedAt":"2026-06-09T03:33:53.819Z","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 3

1.8.8 Integration by substitution (A-level only) Questions

1.8.8 Integration by substitution (A-level only)

Question 3

1.8.8 Integration by substitution (A-level only) Questions