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bg-background","children":["$","$L39",null,{"backHref":"/a-level/maths/wjec/questions","basePath":"/a-level/maths/wjec/integration-10/questions","formulaUrl":"https://www.wjec.co.uk/media/4jfdytcz/ss-w27-18-e.pdf","initialIndex":11,"initialQuestionWindow":[{"id":"e9e83363-2d39-4903-8bb7-db90859b3c62","revisionId":"403d7f15-9545-423d-9041-6a25fa5592a4","questionNumber":null,"questionType":"LA","specification":"","difficulty":3,"difficultyBand":"hard","tier":null,"totalMarks":8,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"9d5e1797-96c1-4943-8c66-30c624ae04c2","label":"a","content":"Find the exact value of $\\int_3^4 \\frac{3x^2 - 3x + 4}{(x-2)^2(2x+1)} dx$","markscheme":"- Sets up partial fractions: $\\frac{A}{x-2} + \\frac{B}{(x-2)^2} + \\frac{C}{2x+1}$ M1\n- Finds constants: $A=1, B=2, C=1$ A1 A1\n- Integrates: $\\left[ \\ln|x-2| - 2(x-2)^{-1} + \\frac{1}{2}\\ln|2x+1| \\right]_3^4$ M1 A1\n- Substitutes limits: $(\\ln 2 - 1 + \\frac{1}{2}\\ln 9) - (0 - 2 + \\frac{1}{2}\\ln 7)$ M1\n- Simplifies: $\\ln 2 - 1 + \\ln 3 + 2 - \\frac{1}{2}\\ln 7$ M1\n- Final exact value: $\\ln 6 - \\frac{1}{2}\\ln 7 + 1$ A1","marks":8,"order":0,"markingRubric":{"totalMarks":8,"points":[{"id":"p1","marks":1,"description":"Sets up the partial fraction decomposition in the form $\\frac{A}{x-2} + \\frac{B}{(x-2)^2} + \\frac{C}{2x+1}$","code":"M"},{"id":"p2","marks":1,"description":"Finds at least two of the constants correctly","code":"A"},{"id":"p3","marks":1,"description":"Correctly finds all three constants $A=1, B=2, C=1$","code":"A"},{"id":"p4","marks":1,"description":"Integration attempt involving natural logs and a power rule term (x-2)^-1","code":"M"},{"id":"p5","marks":1,"description":"Correct integration of their partial fractions, including the 1/2 factor for ln|2x+1|","code":"A"},{"id":"p6","marks":1,"description":"Substitutes limits 3 and 4 into their integrated expression","code":"M"},{"id":"p7","marks":1,"description":"Applies log laws or numerical simplification correctly to at least two terms","code":"M"},{"id":"p8","marks":1,"description":"Correct final exact answer in simplified form","code":"A"}],"finalAnswers":["\\ln 6 - \\frac{1}{2}\\ln 7 + 1","\\ln \\frac{6}{\\sqrt{7}} + 1","\\ln 6 - \\ln \\sqrt{7} + 1"],"generatedAt":"2026-05-06T14:04:46.663Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"direct_partial_fractions","label":"Direct Partial Fractions","steps":[{"type":"text","order":0,"title":"Identify the approach","content":"The objective is to find the exact value of the definite integral:\n\n$$\\int_3^4 \\frac{3x^2 - 3x + 4}{(x-2)^2(2x+1)} \\, dx$$\n\nSince the denominator contains a repeated linear factor $(x-2)^2$ and a distinct linear factor $(2x+1)$, we decompose the integrand directly using partial fractions of the form:\n\n$$\\frac{3x^2 - 3x + 4}{(x-2)^2(2x+1)} \\equiv \\frac{A}{x-2} + \\frac{B}{(x-2)^2} + \\frac{C}{2x+1}$$\n\nThis will allow us to integrate each term individually.","highlights":[{"text":"\\int_3^4 \\frac{3x^2 - 3x + 4}{(x-2)^2(2x+1)} dx","location":"specification"}]},{"type":"text","order":1,"title":"Form the identity","content":"Multiplying both sides of our partial fraction equation by the common denominator $(x-2)^2(2x+1)$, we obtain the following algebraic identity:\n\n$$3x^2 - 3x + 4 \\equiv A(x-2)(2x+1) + B(2x+1) + C(x-2)^2$$\n\nWe can find the constants $A$, $B$, and $C$ by substituting convenient values of $x$ into this identity."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$$3x^2 - 3x + 4 \\equiv A(x-2)(2x+1) + B(2x+1) + C(x-2)^2$$"},{"text":"To find $B$, let $x = 2$:"},{"latex":"$$$3(2)^2 - 3(2) + 4 = B(2(2)+1) \\implies 10 = 5\\htmlData{anchor=b-const}{\\color{@sky}{B}} \\implies \\htmlData{anchor=b-val}{\\color{@sky}{B = 2}}$$"},{"text":"To find $C$, let $x = -\\frac{1}{2}$:"},{"latex":"$$$3\\left(-\\frac{1}{2}\\right)^2 - 3\\left(-\\frac{1}{2}\\right) + 4 = C\\left(-\\frac{1}{2}-2\\right)^2 \\implies \\frac{25}{4} = \\frac{25}{4}\\htmlData{anchor=c-const}{\\color{@amber}{C}} \\implies \\htmlData{anchor=c-val}{\\color{@amber}{C = 1}}$$"},{"text":"To find $A$, equate the coefficients of $x^2$:"},{"latex":"$$$3 = 2\\htmlData{anchor=a-const}{\\color{@purple}{A}} + \\htmlData{anchor=c-coef}{\\color{@amber}{C}} \\implies 3 = 2\\htmlData{anchor=a-solve}{\\color{@purple}{A}} + \\color{@amber}{1} \\implies \\htmlData{anchor=a-val}{\\color{@purple}{A = 1}}$$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"b-val"},{"color":"amber","style":"text-color","anchor":"c-val"},{"color":"purple","style":"text-color","anchor":"a-val"}],"connections":[{"to":"b-const","from":"b-val","color":"sky","label":null,"style":"solid"},{"to":"c-const","from":"c-val","color":"amber","label":null,"style":"solid"},{"to":"c-coef","from":"c-val","color":"amber","label":null,"style":"solid"},{"to":"a-solve","from":"a-val","color":"purple","label":null,"style":"solid"}]},"order":2,"title":"Solve for the constants","description":"We substitute strategic values for $x$ and compare coefficients to find our constants $A = 1$, $B = 2$, and $C = 1$."},{"type":"text","order":3,"title":"Rewrite the integral","content":"Substituting our values $A=1$, $B=2$, and $C=1$ back into the partial fraction decomposition, our integral becomes:\n\n$$\\int_3^4 \\left( \\frac{1}{x-2} + \\frac{2}{(x-2)^2} + \\frac{1}{2x+1} \\right) dx$$"},{"type":"text","order":4,"title":"Integrate term-by-term","content":"Now we integrate each term using standard integration rules:\n- $$\\int \\frac{1}{x-2} \\, dx = \\ln|x-2|$$\n- $$\\int \\frac{2}{(x-2)^2} \\, dx = \\int 2(x-2)^{-2} \\, dx = -2(x-2)^{-1}$$\n- $$\\int \\frac{1}{2x+1} \\, dx = \\frac{1}{2}\\ln|2x+1|$$\n\nCombining these, we get:\n\n$$\\left[ \\ln|x-2| - \\frac{2}{x-2} + \\frac{1}{2}\\ln|2x+1| \\right]_3^4$$"},{"type":"text","order":5,"title":"Evaluate at the boundaries","content":"Next, we evaluate the expression at the upper limit $x=4$ and the lower limit $x=3$:\n\n**At the upper limit ($x=4$):**\n$$\\left( \\ln(2) - \\frac{2}{2} + \\frac{1}{2}\\ln(9) \\right) = \\ln 2 - 1 + \\frac{1}{2}\\ln(3^2) = \\ln 2 - 1 + \\ln 3$$\n\n**At the lower limit ($x=3$):**\n$$\\left( \\ln(1) - \\frac{2}{1} + \\frac{1}{2}\\ln(7) \\right) = 0 - 2 + \\frac{1}{2}\\ln 7 = -2 + \\frac{1}{2}\\ln 7$$"},{"type":"text","order":6,"title":"Subtract and simplify","content":"Subtracting the lower limit evaluation from the upper limit evaluation, we get:\n\n$$\\begin{aligned} \\text{Exact Value} &= (\\ln 2 + \\ln 3 - 1) - \\left(-2 + \\frac{1}{2}\\ln 7\\right) \\\\ &= \\ln 2 + \\ln 3 - 1 + 2 - \\frac{1}{2}\\ln 7 \\\\ &= (\\ln 2 + \\ln 3) - \\frac{1}{2}\\ln 7 + 1 \\end{aligned}$$\n\nUsing log laws, we simplify $\\ln 2 + \\ln 3 = \\ln(2 \\times 3) = \\ln 6$:\n\n$$\\ln 6 - \\frac{1}{2}\\ln 7 + 1$$"},{"type":"text","order":7,"title":"Verify with a GDC","content":"To ensure accuracy, you can use your graphing calculator to numerically check the answer.","calculator":[{"gdc":{"expression":"\\int_{3}^{4}\\frac{3x^2-3x+4}{(x-2)^2(2x+1)}dx","instructions":"1. Go to Run-Matrix mode.\n2. Press [OPTION], then [F4] (CALC), and select [F4] (d/dx template for integration).\n3. Enter the limits 3 and 4, and the integrand expression.\n4. Press [EXE] to get the decimal approximation.\n5. Evaluate ln(6) - 0.5*ln(7) + 1 to check for agreement."},"ti84":{"expression":"\\int_{3}^{4}\\frac{3x^2-3x+4}{(x-2)^2(2x+1)}dx","instructions":"1. Press [MATH] and select option 9 (fnInt).\n2. Enter the integral limits and expression:\n fnInt((3X² - 3X + 4) / ((X - 2)² * (2X + 1)), X, 3, 4)\n3. Press [ENTER] to compute: approximately 1.8159.\n4. Type in your analytical result: ln(6) - 0.5*ln(7) + 1\n5. Press [ENTER] to confirm they match."},"desmos":{"expression":"\\int_{3}^{4}\\frac{3x^{2}-3x+4}{\\left(x-2\\right)^{2}\\left(2x+1\\right)}dx","instructions":"1. Type 'int' to insert an integral template.\n2. Enter 3 in the lower limit box and 4 in the upper limit box.\n3. Type the expression: (3x^2 - 3x + 4)/((x - 2)^2 * (2x + 1))\n4. Compare the result (~1.8159) with ln(6) - 0.5*ln(7) + 1."},"description":"Evaluate the definite integral and the analytical expression to confirm they have the same decimal value."}]}]},{"id":"substitution_partial_fractions","label":"Substitution and Partial Fractions","steps":[{"type":"text","order":0,"title":"Apply linear substitution","content":"To simplify the denominator before decomposing into partial fractions, we use the linear substitution:\n\n$$u = x - 2$$\n\nThis gives:\n- $x = u + 2$\n- $dx = du$\n\nNext, let's convert the integration limits from $x$ to $u$:\n- Lower limit: when $x = 3$, $u = 3 - 2 = 1$\n- Upper limit: when $x = 4$, $u = 4 - 2 = 2$"},{"type":"text","order":1,"title":"Rewrite the integrand","content":"Now we substitute $x = u + 2$ into the numerator and denominator of our integrand.\n\n**Numerator:**\n$$\\begin{aligned}\n3x^2 - 3x + 4 &= 3(u+2)^2 - 3(u+2) + 4 \\\\\n&= 3(u^2 + 4u + 4) - 3u - 6 + 4 \\\\\n&= 3u^2 + 12u + 12 - 3u - 2 \\\\\n&= 3u^2 + 9u + 10\n\\end{aligned}$$\n\n**Denominator:**\n$$(x-2)^2(2x+1) = u^2(2(u+2)+1) = u^2(2u+5)$$\n\nThus, our integral is rewritten in terms of $u$ as:\n$$\\int_1^2 \\frac{3u^2 + 9u + 10}{u^2(2u+5)} du$$"},{"type":"text","order":2,"title":"Set up partial fractions","content":"We can now decompose the new integrand using partial fractions with a repeated linear factor:\n\n$$\\frac{3u^2 + 9u + 10}{u^2(2u+5)} = \\frac{A}{u} + \\frac{B}{u^2} + \\frac{C}{2u+5}$$\n\nMultiplying both sides by the denominator $u^2(2u+5)$ yields the identity:\n\n$$3u^2 + 9u + 10 = Au(2u+5) + B(2u+5) + Cu^2$$"},{"type":"text","order":3,"title":"Find the constants","content":"Let's find the constants $A$, $B$, and $C$ by substituting convenient values for $u$:\n\n- Let $u = 0$:\n $$10 = B(5) \\implies B = 2$$\n\n- Let $u = -\\frac{5}{2}$:\n $$\\begin{aligned}\n 3\\left(-\\frac{5}{2}\\right)^2 + 9\\left(-\\frac{5}{2}\\right) + 10 &= C\\left(-\\frac{5}{2}\\right)^2 \\\\\n 3\\left(\\frac{25}{4}\\right) - \\frac{45}{2} + 10 &= \\frac{25}{4}C \\\\\n \\frac{75}{4} - \\frac{90}{4} + \\frac{40}{4} &= \\frac{25}{4}C \\\\\n \\frac{25}{4} &= \\frac{25}{4}C \\implies C = 1\n \\end{aligned}$$\n\n- To find $A$, compare coefficients of $u^2$ on both sides:\n $$3 = 2A + C$$\n Substitute $C=1$:\n $$3 = 2A + 1 \\implies 2A = 2 \\implies A = 1$$\n\nSo, our partial fraction decomposition is:\n$$\\frac{3u^2 + 9u + 10}{u^2(2u+5)} = \\frac{1}{u} + \\frac{2}{u^2} + \\frac{1}{2u+5}$$\n\n*(Note: This is equivalent to the original decomposition $\\frac{1}{x-2} + \\frac{2}{(x-2)^2} + \\frac{1}{2x+1}$ but simplified under our substitution).* "},{"type":"text","order":4,"title":"Integrate the terms","content":"Now we integrate each term with respect to $u$ from $1$ to $2$:\n\n$$\\int_1^2 \\left( \\frac{1}{u} + 2u^{-2} + \\frac{1}{2u+5} \\right) du$$\n\nUsing standard integration rules:\n$$\\int \\frac{1}{u} du = \\ln|u|$$\n$$\\int 2u^{-2} du = -\\frac{2}{u}$$\n$$\\int \\frac{1}{2u+5} du = \\frac{1}{2}\\ln|2u+5|$$\n\nThis gives us the evaluated antiderivative:\n$$\\left[ \\ln|u| - \\frac{2}{u} + \\frac{1}{2}\\ln|2u+5| \\right]_1^2$$"},{"type":"text","order":5,"title":"Evaluate at the limits","content":"Substitute the upper limit $u = 2$ and the lower limit $u = 1$:\n\n- **At $u = 2$**:\n $$\\ln(2) - \\frac{2}{2} + \\frac{1}{2}\\ln(2(2)+5) = \\ln 2 - 1 + \\frac{1}{2}\\ln 9$$\n\n- **At $u = 1$**:\n $$\\ln(1) - \\frac{2}{1} + \\frac{1}{2}\\ln(2(1)+5) = 0 - 2 + \\frac{1}{2}\\ln 7$$\n\nSubtract the lower limit evaluation from the upper limit evaluation:\n$$\\left( \\ln 2 - 1 + \\frac{1}{2}\\ln 9 \\right) - \\left( -2 + \\frac{1}{2}\\ln 7 \\right)$$"},{"type":"text","order":6,"title":"Simplify to final exact value","content":"Let's simplify the expression using logarithmic properties:\n\n- Note that $\\frac{1}{2}\\ln 9 = \\ln(9^{1/2}) = \\ln 3$\n\nNow substitute this back and simplify the constant terms:\n$$\\begin{aligned}\n\\text{Value} &= \\ln 2 - 1 + \\ln 3 + 2 - \\frac{1}{2}\\ln 7 \\\\\n&= (\\ln 2 + \\ln 3) + 1 - \\frac{1}{2}\\ln 7 \\\\\n&= \\ln(2 \\times 3) - \\frac{1}{2}\\ln 7 + 1 \\\\\n&= \\ln 6 - \\frac{1}{2}\\ln 7 + 1\n\\end{aligned}$$\n\n**Final exact value:**\n$$\\ln 6 - \\frac{1}{2}\\ln 7 + 1$$"}]}],"generatedAt":"2026-06-11T03:23:53.144Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"9d5e1797-96c1-4943-8c66-30c624ae04c2:direct_partial_fractions","label":"Direct Partial Fractions","steps":[{"type":"text","order":0,"title":"Identify the approach","content":"The objective is to find the exact value of the definite integral:\n\n$$\\int_3^4 \\frac{3x^2 - 3x + 4}{(x-2)^2(2x+1)} \\, dx$$\n\nSince the denominator contains a repeated linear factor $(x-2)^2$ and a distinct linear factor $(2x+1)$, we decompose the integrand directly using partial fractions of the form:\n\n$$\\frac{3x^2 - 3x + 4}{(x-2)^2(2x+1)} \\equiv \\frac{A}{x-2} + \\frac{B}{(x-2)^2} + \\frac{C}{2x+1}$$\n\nThis will allow us to integrate each term individually.","highlights":[{"text":"\\int_3^4 \\frac{3x^2 - 3x + 4}{(x-2)^2(2x+1)} dx","location":"specification"}]},{"type":"text","order":1,"title":"Form the identity","content":"Multiplying both sides of our partial fraction equation by the common denominator $(x-2)^2(2x+1)$, we obtain the following algebraic identity:\n\n$$3x^2 - 3x + 4 \\equiv A(x-2)(2x+1) + B(2x+1) + C(x-2)^2$$\n\nWe can find the constants $A$, $B$, and $C$ by substituting convenient values of $x$ into this identity."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$$3x^2 - 3x + 4 \\equiv A(x-2)(2x+1) + B(2x+1) + C(x-2)^2$$"},{"text":"To find $B$, let $x = 2$:"},{"latex":"$$$3(2)^2 - 3(2) + 4 = B(2(2)+1) \\implies 10 = 5\\htmlData{anchor=b-const}{\\color{@sky}{B}} \\implies \\htmlData{anchor=b-val}{\\color{@sky}{B = 2}}$$"},{"text":"To find $C$, let $x = -\\frac{1}{2}$:"},{"latex":"$$$3\\left(-\\frac{1}{2}\\right)^2 - 3\\left(-\\frac{1}{2}\\right) + 4 = C\\left(-\\frac{1}{2}-2\\right)^2 \\implies \\frac{25}{4} = \\frac{25}{4}\\htmlData{anchor=c-const}{\\color{@amber}{C}} \\implies \\htmlData{anchor=c-val}{\\color{@amber}{C = 1}}$$"},{"text":"To find $A$, equate the coefficients of $x^2$:"},{"latex":"$$$3 = 2\\htmlData{anchor=a-const}{\\color{@purple}{A}} + \\htmlData{anchor=c-coef}{\\color{@amber}{C}} \\implies 3 = 2\\htmlData{anchor=a-solve}{\\color{@purple}{A}} + \\color{@amber}{1} \\implies \\htmlData{anchor=a-val}{\\color{@purple}{A = 1}}$$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"b-val"},{"color":"amber","style":"text-color","anchor":"c-val"},{"color":"purple","style":"text-color","anchor":"a-val"}],"connections":[{"to":"b-const","from":"b-val","color":"sky","label":null,"style":"solid"},{"to":"c-const","from":"c-val","color":"amber","label":null,"style":"solid"},{"to":"c-coef","from":"c-val","color":"amber","label":null,"style":"solid"},{"to":"a-solve","from":"a-val","color":"purple","label":null,"style":"solid"}]},"order":2,"title":"Solve for the constants","description":"We substitute strategic values for $x$ and compare coefficients to find our constants $A = 1$, $B = 2$, and $C = 1$."},{"type":"text","order":3,"title":"Rewrite the integral","content":"Substituting our values $A=1$, $B=2$, and $C=1$ back into the partial fraction decomposition, our integral becomes:\n\n$$\\int_3^4 \\left( \\frac{1}{x-2} + \\frac{2}{(x-2)^2} + \\frac{1}{2x+1} \\right) dx$$"},{"type":"text","order":4,"title":"Integrate term-by-term","content":"Now we integrate each term using standard integration rules:\n- $$\\int \\frac{1}{x-2} \\, dx = \\ln|x-2|$$\n- $$\\int \\frac{2}{(x-2)^2} \\, dx = \\int 2(x-2)^{-2} \\, dx = -2(x-2)^{-1}$$\n- $$\\int \\frac{1}{2x+1} \\, dx = \\frac{1}{2}\\ln|2x+1|$$\n\nCombining these, we get:\n\n$$\\left[ \\ln|x-2| - \\frac{2}{x-2} + \\frac{1}{2}\\ln|2x+1| \\right]_3^4$$"},{"type":"text","order":5,"title":"Evaluate at the boundaries","content":"Next, we evaluate the expression at the upper limit $x=4$ and the lower limit $x=3$:\n\n**At the upper limit ($x=4$):**\n$$\\left( \\ln(2) - \\frac{2}{2} + \\frac{1}{2}\\ln(9) \\right) = \\ln 2 - 1 + \\frac{1}{2}\\ln(3^2) = \\ln 2 - 1 + \\ln 3$$\n\n**At the lower limit ($x=3$):**\n$$\\left( \\ln(1) - \\frac{2}{1} + \\frac{1}{2}\\ln(7) \\right) = 0 - 2 + \\frac{1}{2}\\ln 7 = -2 + \\frac{1}{2}\\ln 7$$"},{"type":"text","order":6,"title":"Subtract and simplify","content":"Subtracting the lower limit evaluation from the upper limit evaluation, we get:\n\n$$\\begin{aligned} \\text{Exact Value} &= (\\ln 2 + \\ln 3 - 1) - \\left(-2 + \\frac{1}{2}\\ln 7\\right) \\\\ &= \\ln 2 + \\ln 3 - 1 + 2 - \\frac{1}{2}\\ln 7 \\\\ &= (\\ln 2 + \\ln 3) - \\frac{1}{2}\\ln 7 + 1 \\end{aligned}$$\n\nUsing log laws, we simplify $\\ln 2 + \\ln 3 = \\ln(2 \\times 3) = \\ln 6$:\n\n$$\\ln 6 - \\frac{1}{2}\\ln 7 + 1$$"},{"type":"text","order":7,"title":"Verify with a GDC","content":"To ensure accuracy, you can use your graphing calculator to numerically check the answer.","calculator":[{"gdc":{"expression":"\\int_{3}^{4}\\frac{3x^2-3x+4}{(x-2)^2(2x+1)}dx","instructions":"1. Go to Run-Matrix mode.\n2. Press [OPTION], then [F4] (CALC), and select [F4] (d/dx template for integration).\n3. Enter the limits 3 and 4, and the integrand expression.\n4. Press [EXE] to get the decimal approximation.\n5. Evaluate ln(6) - 0.5*ln(7) + 1 to check for agreement."},"ti84":{"expression":"\\int_{3}^{4}\\frac{3x^2-3x+4}{(x-2)^2(2x+1)}dx","instructions":"1. Press [MATH] and select option 9 (fnInt).\n2. Enter the integral limits and expression:\n fnInt((3X² - 3X + 4) / ((X - 2)² * (2X + 1)), X, 3, 4)\n3. Press [ENTER] to compute: approximately 1.8159.\n4. Type in your analytical result: ln(6) - 0.5*ln(7) + 1\n5. Press [ENTER] to confirm they match."},"desmos":{"expression":"\\int_{3}^{4}\\frac{3x^{2}-3x+4}{\\left(x-2\\right)^{2}\\left(2x+1\\right)}dx","instructions":"1. Type 'int' to insert an integral template.\n2. Enter 3 in the lower limit box and 4 in the upper limit box.\n3. Type the expression: (3x^2 - 3x + 4)/((x - 2)^2 * (2x + 1))\n4. Compare the result (~1.8159) with ln(6) - 0.5*ln(7) + 1."},"description":"Evaluate the definite integral and the analytical expression to confirm they have the same decimal value."}]}]},{"id":"9d5e1797-96c1-4943-8c66-30c624ae04c2:substitution_partial_fractions","label":"Substitution and Partial Fractions","steps":[{"type":"text","order":0,"title":"Apply linear substitution","content":"To simplify the denominator before decomposing into partial fractions, we use the linear substitution:\n\n$$u = x - 2$$\n\nThis gives:\n- $x = u + 2$\n- $dx = du$\n\nNext, let's convert the integration limits from $x$ to $u$:\n- Lower limit: when $x = 3$, $u = 3 - 2 = 1$\n- Upper limit: when $x = 4$, $u = 4 - 2 = 2$"},{"type":"text","order":1,"title":"Rewrite the integrand","content":"Now we substitute $x = u + 2$ into the numerator and denominator of our integrand.\n\n**Numerator:**\n$$\\begin{aligned}\n3x^2 - 3x + 4 &= 3(u+2)^2 - 3(u+2) + 4 \\\\\n&= 3(u^2 + 4u + 4) - 3u - 6 + 4 \\\\\n&= 3u^2 + 12u + 12 - 3u - 2 \\\\\n&= 3u^2 + 9u + 10\n\\end{aligned}$$\n\n**Denominator:**\n$$(x-2)^2(2x+1) = u^2(2(u+2)+1) = u^2(2u+5)$$\n\nThus, our integral is rewritten in terms of $u$ as:\n$$\\int_1^2 \\frac{3u^2 + 9u + 10}{u^2(2u+5)} du$$"},{"type":"text","order":2,"title":"Set up partial fractions","content":"We can now decompose the new integrand using partial fractions with a repeated linear factor:\n\n$$\\frac{3u^2 + 9u + 10}{u^2(2u+5)} = \\frac{A}{u} + \\frac{B}{u^2} + \\frac{C}{2u+5}$$\n\nMultiplying both sides by the denominator $u^2(2u+5)$ yields the identity:\n\n$$3u^2 + 9u + 10 = Au(2u+5) + B(2u+5) + Cu^2$$"},{"type":"text","order":3,"title":"Find the constants","content":"Let's find the constants $A$, $B$, and $C$ by substituting convenient values for $u$:\n\n- Let $u = 0$:\n $$10 = B(5) \\implies B = 2$$\n\n- Let $u = -\\frac{5}{2}$:\n $$\\begin{aligned}\n 3\\left(-\\frac{5}{2}\\right)^2 + 9\\left(-\\frac{5}{2}\\right) + 10 &= C\\left(-\\frac{5}{2}\\right)^2 \\\\\n 3\\left(\\frac{25}{4}\\right) - \\frac{45}{2} + 10 &= \\frac{25}{4}C \\\\\n \\frac{75}{4} - \\frac{90}{4} + \\frac{40}{4} &= \\frac{25}{4}C \\\\\n \\frac{25}{4} &= \\frac{25}{4}C \\implies C = 1\n \\end{aligned}$$\n\n- To find $A$, compare coefficients of $u^2$ on both sides:\n $$3 = 2A + C$$\n Substitute $C=1$:\n $$3 = 2A + 1 \\implies 2A = 2 \\implies A = 1$$\n\nSo, our partial fraction decomposition is:\n$$\\frac{3u^2 + 9u + 10}{u^2(2u+5)} = \\frac{1}{u} + \\frac{2}{u^2} + \\frac{1}{2u+5}$$\n\n*(Note: This is equivalent to the original decomposition $\\frac{1}{x-2} + \\frac{2}{(x-2)^2} + \\frac{1}{2x+1}$ but simplified under our substitution).* "},{"type":"text","order":4,"title":"Integrate the terms","content":"Now we integrate each term with respect to $u$ from $1$ to $2$:\n\n$$\\int_1^2 \\left( \\frac{1}{u} + 2u^{-2} + \\frac{1}{2u+5} \\right) du$$\n\nUsing standard integration rules:\n$$\\int \\frac{1}{u} du = \\ln|u|$$\n$$\\int 2u^{-2} du = -\\frac{2}{u}$$\n$$\\int \\frac{1}{2u+5} du = \\frac{1}{2}\\ln|2u+5|$$\n\nThis gives us the evaluated antiderivative:\n$$\\left[ \\ln|u| - \\frac{2}{u} + \\frac{1}{2}\\ln|2u+5| \\right]_1^2$$"},{"type":"text","order":5,"title":"Evaluate at the limits","content":"Substitute the upper limit $u = 2$ and the lower limit $u = 1$:\n\n- **At $u = 2$**:\n $$\\ln(2) - \\frac{2}{2} + \\frac{1}{2}\\ln(2(2)+5) = \\ln 2 - 1 + \\frac{1}{2}\\ln 9$$\n\n- **At $u = 1$**:\n $$\\ln(1) - \\frac{2}{1} + \\frac{1}{2}\\ln(2(1)+5) = 0 - 2 + \\frac{1}{2}\\ln 7$$\n\nSubtract the lower limit evaluation from the upper limit evaluation:\n$$\\left( \\ln 2 - 1 + \\frac{1}{2}\\ln 9 \\right) - \\left( -2 + \\frac{1}{2}\\ln 7 \\right)$$"},{"type":"text","order":6,"title":"Simplify to final exact value","content":"Let's simplify the expression using logarithmic properties:\n\n- Note that $\\frac{1}{2}\\ln 9 = \\ln(9^{1/2}) = \\ln 3$\n\nNow substitute this back and simplify the constant terms:\n$$\\begin{aligned}\n\\text{Value} &= \\ln 2 - 1 + \\ln 3 + 2 - \\frac{1}{2}\\ln 7 \\\\\n&= (\\ln 2 + \\ln 3) + 1 - \\frac{1}{2}\\ln 7 \\\\\n&= \\ln(2 \\times 3) - \\frac{1}{2}\\ln 7 + 1 \\\\\n&= \\ln 6 - \\frac{1}{2}\\ln 7 + 1\n\\end{aligned}$$\n\n**Final exact value:**\n$$\\ln 6 - \\frac{1}{2}\\ln 7 + 1$$"}]}],"generatedAt":"2026-06-11T03:23:53.200Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null},{"id":"0cba29e6-3d13-4501-a510-cb4d1b065743","revisionId":"08ce8717-2090-4d51-b50a-d6b26b868bcd","questionNumber":"7","questionType":"LA","specification":"","difficulty":3,"difficultyBand":"hard","tier":null,"totalMarks":8,"sourceTitle":"Integration","calculatorUse":null,"isCalculation":true,"parts":[{"id":"c20cd5ac-305b-49d6-91b3-82fb970a8ece","label":"0","content":"Find the exact value of $\\int_0^5 \\frac{2x^2 + 3x + 7}{(x + 1)^2(x + 3)} dx$","markscheme":"- Sets up partial fractions: $\\frac{A}{x+1} + \\frac{B}{(x+1)^2} + \\frac{C}{x+3}$ M1\n- Finds constants: $B=3, C=4, A=-2$ A1 A1\n- Integrates: $\\left[ -3(x+1)^{-1} + 4\\ln|x+3| - 2\\ln|x+1| \\right]_0^5$ M1 A1\n- Substitutes limits: $\\left( -\\frac{1}{2} + 4\\ln 8 - 2\\ln 6 \\right) - (-3 + 4\\ln 3)$ M1\n- Simplifies: $\\frac{5}{2} + 12\\ln 2 - 2\\ln 6 - 4\\ln 3$ M1\n- Final exact value A1","marks":8,"order":0,"markingRubric":{"totalMarks":8,"points":[{"id":"p1","marks":1,"description":"Correct form for partial fractions with repeated factor.","code":"M"},{"id":"p2","marks":1,"description":"Correctly finds constants $B=3$ and $C=4$.","code":"A"},{"id":"p3","marks":1,"description":"Correctly finds constant $A=-2$.","code":"A"},{"id":"p4","marks":1,"description":"Integrates $\\frac{B}{(x+1)^2}$ to $-3(x+1)^{-1}$.","code":"M"},{"id":"p5","marks":1,"description":"Integrates log terms correctly: $4\\ln|x+3| - 2\\ln|x+1|$.","code":"A"},{"id":"p6","marks":1,"description":"Substitutes limits 5 and 0 into the integrated expression.","code":"M"},{"id":"p7","marks":1,"description":"Combines constant terms to $\\frac{5}{2}$.","code":"A"},{"id":"p8","marks":1,"description":"Simplifies log terms to an exact form.","code":"A"}],"finalAnswers":["\\frac{5}{2} + 12\\ln 2 - 2\\ln 6 - 4\\ln 3"],"generatedAt":"2026-05-06T03:37:59.137Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"direct_partial_fractions","label":"Direct Partial Fractions","steps":[{"type":"text","order":0,"title":"Set up partial fractions","content":"To integrate the rational function, we first decompose the integrand into partial fractions. Because we have a repeated linear factor $(x+1)^2$ and a distinct linear factor $(x+3)$ in the denominator, the decomposition takes the form:\n\n$$\\frac{2x^2 + 3x + 7}{(x+1)^2(x+3)} = \\frac{A}{x+1} + \\frac{B}{(x+1)^2} + \\frac{C}{x+3}$$"},{"type":"text","order":1,"title":"Find the constants","content":"Multiply both sides of the equation by the common denominator $(x+1)^2(x+3)$ to clear the fractions:\n\n$$2x^2 + 3x + 7 = A(x+1)(x+3) + B(x+3) + C(x+1)^2$$\n\nNow we can find the constants $A$, $B$, and $C$ by substituting convenient values for $x$:\n\n- **Set $x = -1$**:\n$$2(-1)^2 + 3(-1) + 7 = B(-1+3)$$\n$$2 - 3 + 7 = 2B$$\n$$6 = 2B \\implies B = 3$$\n\n- **Set $x = -3$**:\n$$2(-3)^2 + 3(-3) + 7 = C(-3+1)^2$$\n$$18 - 9 + 7 = 4C$$\n$$16 = 4C \\implies C = 4$$\n\n- **Compare coefficients of $x^2$**:\n$$2 = A + C$$\n\nSince $C = 4$:\n$$2 = A + 4 \\implies A = -2$$\n\nSo, our partial fraction decomposition is:\n\n$$\\frac{2x^2 + 3x + 7}{(x+1)^2(x+3)} = -\\frac{2}{x+1} + \\frac{3}{(x+1)^2} + \\frac{4}{x+3}$$"},{"type":"text","order":2,"title":"Integrate each term","content":"Now we substitute the partial fraction decomposition back into the integral and integrate term by term:\n\n$$\\int_0^5 \\left( -\\frac{2}{x+1} + 3(x+1)^{-2} + \\frac{4}{x+3} \\right) dx$$\n\nUsing standard integration rules, we get:\n\n$$\\int -\\frac{2}{x+1} dx = -2\\ln|x+1|$$\n\n$$\\int 3(x+1)^{-2} dx = -3(x+1)^{-1} = -\\frac{3}{x+1}$$\n\n$$\\int \\frac{4}{x+3} dx = 4\\ln|x+3|$$\n\nCombining these, the integrated expression is:\n\n$$\\left[ -2\\ln|x+1| - \\frac{3}{x+1} + 4\\ln|x+3| \\right]_0^5$$"},{"type":"text","order":3,"title":"Apply limits of integration","content":"Now we substitute the upper limit $x = 5$ and the lower limit $x = 0$:\n\n- **At $x = 5$**:\n$$-2\\ln(6) - \\frac{3}{6} + 4\\ln(8) = -2\\ln 6 - \\frac{1}{2} + 4\\ln 8$$\n\n- **At $x = 0$**:\n$$-2\\ln(1) - \\frac{3}{1} + 4\\ln(3) = 0 - 3 + 4\\ln 3 = -3 + 4\\ln 3$$\n\nSubtract the lower limit evaluated expression from the upper limit evaluated expression:\n\n$$\\left( -\\frac{1}{2} - 2\\ln 6 + 4\\ln 8 \\right) - \\left( -3 + 4\\ln 3 \\right)$$\n\nGroup the constant terms and the logarithm terms:\n\n$$\\left(-\\frac{1}{2} + 3\\right) + 4\\ln 8 - 2\\ln 6 - 4\\ln 3 = \\frac{5}{2} + 4\\ln(2^3) - 2\\ln(2 \\times 3) - 4\\ln 3$$"},{"type":"text","order":4,"title":"Simplify the exact value","content":"Let's simplify the logarithm terms to express the final answer in terms of prime factors:\n\n$$4\\ln 8 = 4\\ln(2^3) = 12\\ln 2$$\n\n$$2\\ln 6 = 2(\\ln 2 + \\ln 3) = 2\\ln 2 + 2\\ln 3$$\n\nSubstitute these back into our expression:\n\n$$\\frac{5}{2} + 12\\ln 2 - (2\\ln 2 + 2\\ln 3) - 4\\ln 3 = \\frac{5}{2} + 12\\ln 2 - 2\\ln 2 - 2\\ln 3 - 4\\ln 3$$\n\n$$\\frac{5}{2} + 12\\ln 2 - 2\\ln 2 - 2\\ln 3 - 4\\ln 3 = \\frac{5}{2} + 10\\ln 2 - 6\\ln 3$$\n\nThis can also be written in other equivalent forms, such as:\n\n$$\\frac{5}{2} + \\ln\\left(\\frac{1024}{729}\\right)$$\n\nThus, the exact value of the integral is:\n\n$$\\frac{5}{2} + 10\\ln 2 - 6\\ln 3$$"}]},{"id":"substitution_then_partial_fractions","label":"Substitution then Partial Fractions","steps":[{"type":"text","order":0,"title":"Introduce the substitution","content":"To simplify the denominator of the integrand, we will perform a linear substitution first.\n\nLet's choose:\n$$u = x + 1$$\n\nFrom this, we get:\n$$x = u - 1$$\n\n$$dx = du$$\n\nWe also need to transform the integration limits:\n- When $x = 0$, $u = 0 + 1 = 1$\n- When $x = 5$, $u = 5 + 1 = 6$"},{"type":"text","order":1,"title":"Substitute into the integrand","content":"Now we express the numerator and denominator in terms of $u$:\n- Denominator: $(x+1)^2(x+3) = u^2(u + 2)$\n- Numerator:\n$$2x^2 + 3x + 7 = 2(u-1)^2 + 3(u-1) + 7$$\n$$2x^2 + 3x + 7 = 2(u^2 - 2u + 1) + 3u - 3 + 7$$\n$$2x^2 + 3x + 7 = 2u^2 - 4u + 2 + 3u + 4$$\n$$2x^2 + 3x + 7 = 2u^2 - u + 6$$\n\nSubstituting these and the new limits, the integral becomes:\n$$\\int_1^6 \\frac{2u^2 - u + 6}{u^2(u + 2)} du$$"},{"type":"text","order":2,"title":"Formulate partial fractions","content":"Next, we decompose the integrand into partial fractions:\n$$\\frac{2u^2 - u + 6}{u^2(u + 2)} = \\frac{A}{u} + \\frac{B}{u^2} + \\frac{C}{u + 2}$$\n\nMultiply both sides by the common denominator $u^2(u + 2)$ to clear the fractions:\n$$2u^2 - u + 6 = Au(u + 2) + B(u + 2) + Cu^2$$"},{"type":"text","order":3,"title":"Determine the constants","content":"We can find the constants $A$, $B$, and $C$ by substituting convenient values for $u$:\n\n**To find $B$**:\nLet $u = 0$:\n$$2(0)^2 - 0 + 6 = B(0 + 2)$$\n$$6 = 2B \\implies B = 3$$\n\n**To find $C$**:\nLet $u = -2$:\n$$2(-2)^2 - (-2) + 6 = C(-2)^2$$\n$$8 + 2 + 6 = 4C$$\n$$16 = 4C \\implies C = 4$$\n\n**To find $A$**:\nComparing the coefficients of $u^2$ on both sides:\n$$2 = A + C$$\n$$2 = A + 4 \\implies A = -2$$\n\nSo, the partial fraction decomposition is:\n$$\\frac{2u^2 - u + 6}{u^2(u + 2)} = -\\frac{2}{u} + \\frac{3}{u^2} + \\frac{4}{u + 2}$$\n\n*(Note: These match the coefficients $A=-2$, $B=3$, $C=4$ in the markscheme!)*"},{"type":"text","order":4,"title":"Integrate with transformed limits","content":"Now we substitute our decomposed form back into the integral and integrate term by term:\n$$\\int_1^6 \\left( -2u^{-1} + 3u^{-2} + 4(u+2)^{-1} \\right) du$$\n\nIntegrating each term gives:\n$$\\left[ -2\\ln|u| - \\frac{3}{u} + 4\\ln|u + 2| \\right]_1^6$$"},{"type":"text","order":5,"title":"Evaluate the definite integral","content":"Now, we apply the integration limits from $u = 1$ to $u = 6$:\n\n**Upper limit ($u = 6$):**\n$$-2\\ln 6 - \\frac{3}{6} + 4\\ln 8 = -2\\ln 6 - \\frac{1}{2} + 4\\ln 8$$\n\n**Lower limit ($u = 1$):**\n$$-2\\ln 1 - \\frac{3}{1} + 4\\ln 3 = -3 + 4\\ln 3$$\n\nSubtracting the lower limit term from the upper limit term:\n$$\\left( -\\frac{1}{2} - 2\\ln 6 + 4\\ln 8 \\right) - (-3 + 4\\ln 3)$$\n$$\\left( -\\frac{1}{2} - 2\\ln 6 + 4\\ln 8 \\right) - (-3 + 4\\ln 3) = \\frac{5}{2} + 4\\ln 8 - 2\\ln 6 - 4\\ln 3$$"},{"type":"text","order":6,"title":"Simplify using logarithm laws","content":"Let's simplify the logarithmic expressions to find the final exact value:\n\n- Write $8$ as $2^3$:\n$$4\\ln 8 = 4\\ln(2^3) = 12\\ln 2$$\n- Write $6$ as $2 \\times 3$:\n$$2\\ln 6 = 2(\\ln 2 + \\ln 3) = 2\\ln 2 + 2\\ln 3$$\n\nSubstitute these back into our expression:\n$$\\frac{5}{2} + 12\\ln 2 - (2\\ln 2 + 2\\ln 3) - 4\\ln 3 = \\frac{5}{2} + 12\\ln 2 - 2\\ln 2 - 2\\ln 3 - 4\\ln 3$$\n$$\\frac{5}{2} + 12\\ln 2 - 2\\ln 2 - 2\\ln 3 - 4\\ln 3 = \\frac{5}{2} + 10\\ln 2 - 6\\ln 3$$\n\nThus, the exact value of the integral is:\n$$\\frac{5}{2} + 10\\ln 2 - 6\\ln 3$$"}]}],"generatedAt":"2026-06-09T03:16:21.209Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"c20cd5ac-305b-49d6-91b3-82fb970a8ece:direct_partial_fractions","label":"Direct Partial Fractions","steps":[{"type":"text","order":0,"title":"Set up partial fractions","content":"To integrate the rational function, we first decompose the integrand into partial fractions. Because we have a repeated linear factor $(x+1)^2$ and a distinct linear factor $(x+3)$ in the denominator, the decomposition takes the form:\n\n$$\\frac{2x^2 + 3x + 7}{(x+1)^2(x+3)} = \\frac{A}{x+1} + \\frac{B}{(x+1)^2} + \\frac{C}{x+3}$$"},{"type":"text","order":1,"title":"Find the constants","content":"Multiply both sides of the equation by the common denominator $(x+1)^2(x+3)$ to clear the fractions:\n\n$$2x^2 + 3x + 7 = A(x+1)(x+3) + B(x+3) + C(x+1)^2$$\n\nNow we can find the constants $A$, $B$, and $C$ by substituting convenient values for $x$:\n\n- **Set $x = -1$**:\n$$2(-1)^2 + 3(-1) + 7 = B(-1+3)$$\n$$2 - 3 + 7 = 2B$$\n$$6 = 2B \\implies B = 3$$\n\n- **Set $x = -3$**:\n$$2(-3)^2 + 3(-3) + 7 = C(-3+1)^2$$\n$$18 - 9 + 7 = 4C$$\n$$16 = 4C \\implies C = 4$$\n\n- **Compare coefficients of $x^2$**:\n$$2 = A + C$$\n\nSince $C = 4$:\n$$2 = A + 4 \\implies A = -2$$\n\nSo, our partial fraction decomposition is:\n\n$$\\frac{2x^2 + 3x + 7}{(x+1)^2(x+3)} = -\\frac{2}{x+1} + \\frac{3}{(x+1)^2} + \\frac{4}{x+3}$$"},{"type":"text","order":2,"title":"Integrate each term","content":"Now we substitute the partial fraction decomposition back into the integral and integrate term by term:\n\n$$\\int_0^5 \\left( -\\frac{2}{x+1} + 3(x+1)^{-2} + \\frac{4}{x+3} \\right) dx$$\n\nUsing standard integration rules, we get:\n\n$$\\int -\\frac{2}{x+1} dx = -2\\ln|x+1|$$\n\n$$\\int 3(x+1)^{-2} dx = -3(x+1)^{-1} = -\\frac{3}{x+1}$$\n\n$$\\int \\frac{4}{x+3} dx = 4\\ln|x+3|$$\n\nCombining these, the integrated expression is:\n\n$$\\left[ -2\\ln|x+1| - \\frac{3}{x+1} + 4\\ln|x+3| \\right]_0^5$$"},{"type":"text","order":3,"title":"Apply limits of integration","content":"Now we substitute the upper limit $x = 5$ and the lower limit $x = 0$:\n\n- **At $x = 5$**:\n$$-2\\ln(6) - \\frac{3}{6} + 4\\ln(8) = -2\\ln 6 - \\frac{1}{2} + 4\\ln 8$$\n\n- **At $x = 0$**:\n$$-2\\ln(1) - \\frac{3}{1} + 4\\ln(3) = 0 - 3 + 4\\ln 3 = -3 + 4\\ln 3$$\n\nSubtract the lower limit evaluated expression from the upper limit evaluated expression:\n\n$$\\left( -\\frac{1}{2} - 2\\ln 6 + 4\\ln 8 \\right) - \\left( -3 + 4\\ln 3 \\right)$$\n\nGroup the constant terms and the logarithm terms:\n\n$$\\left(-\\frac{1}{2} + 3\\right) + 4\\ln 8 - 2\\ln 6 - 4\\ln 3 = \\frac{5}{2} + 4\\ln(2^3) - 2\\ln(2 \\times 3) - 4\\ln 3$$"},{"type":"text","order":4,"title":"Simplify the exact value","content":"Let's simplify the logarithm terms to express the final answer in terms of prime factors:\n\n$$4\\ln 8 = 4\\ln(2^3) = 12\\ln 2$$\n\n$$2\\ln 6 = 2(\\ln 2 + \\ln 3) = 2\\ln 2 + 2\\ln 3$$\n\nSubstitute these back into our expression:\n\n$$\\frac{5}{2} + 12\\ln 2 - (2\\ln 2 + 2\\ln 3) - 4\\ln 3 = \\frac{5}{2} + 12\\ln 2 - 2\\ln 2 - 2\\ln 3 - 4\\ln 3$$\n\n$$\\frac{5}{2} + 12\\ln 2 - 2\\ln 2 - 2\\ln 3 - 4\\ln 3 = \\frac{5}{2} + 10\\ln 2 - 6\\ln 3$$\n\nThis can also be written in other equivalent forms, such as:\n\n$$\\frac{5}{2} + \\ln\\left(\\frac{1024}{729}\\right)$$\n\nThus, the exact value of the integral is:\n\n$$\\frac{5}{2} + 10\\ln 2 - 6\\ln 3$$"}]},{"id":"c20cd5ac-305b-49d6-91b3-82fb970a8ece:substitution_then_partial_fractions","label":"Substitution then Partial Fractions","steps":[{"type":"text","order":0,"title":"Introduce the substitution","content":"To simplify the denominator of the integrand, we will perform a linear substitution first.\n\nLet's choose:\n$$u = x + 1$$\n\nFrom this, we get:\n$$x = u - 1$$\n\n$$dx = du$$\n\nWe also need to transform the integration limits:\n- When $x = 0$, $u = 0 + 1 = 1$\n- When $x = 5$, $u = 5 + 1 = 6$"},{"type":"text","order":1,"title":"Substitute into the integrand","content":"Now we express the numerator and denominator in terms of $u$:\n- Denominator: $(x+1)^2(x+3) = u^2(u + 2)$\n- Numerator:\n$$2x^2 + 3x + 7 = 2(u-1)^2 + 3(u-1) + 7$$\n$$2x^2 + 3x + 7 = 2(u^2 - 2u + 1) + 3u - 3 + 7$$\n$$2x^2 + 3x + 7 = 2u^2 - 4u + 2 + 3u + 4$$\n$$2x^2 + 3x + 7 = 2u^2 - u + 6$$\n\nSubstituting these and the new limits, the integral becomes:\n$$\\int_1^6 \\frac{2u^2 - u + 6}{u^2(u + 2)} du$$"},{"type":"text","order":2,"title":"Formulate partial fractions","content":"Next, we decompose the integrand into partial fractions:\n$$\\frac{2u^2 - u + 6}{u^2(u + 2)} = \\frac{A}{u} + \\frac{B}{u^2} + \\frac{C}{u + 2}$$\n\nMultiply both sides by the common denominator $u^2(u + 2)$ to clear the fractions:\n$$2u^2 - u + 6 = Au(u + 2) + B(u + 2) + Cu^2$$"},{"type":"text","order":3,"title":"Determine the constants","content":"We can find the constants $A$, $B$, and $C$ by substituting convenient values for $u$:\n\n**To find $B$**:\nLet $u = 0$:\n$$2(0)^2 - 0 + 6 = B(0 + 2)$$\n$$6 = 2B \\implies B = 3$$\n\n**To find $C$**:\nLet $u = -2$:\n$$2(-2)^2 - (-2) + 6 = C(-2)^2$$\n$$8 + 2 + 6 = 4C$$\n$$16 = 4C \\implies C = 4$$\n\n**To find $A$**:\nComparing the coefficients of $u^2$ on both sides:\n$$2 = A + C$$\n$$2 = A + 4 \\implies A = -2$$\n\nSo, the partial fraction decomposition is:\n$$\\frac{2u^2 - u + 6}{u^2(u + 2)} = -\\frac{2}{u} + \\frac{3}{u^2} + \\frac{4}{u + 2}$$\n\n*(Note: These match the coefficients $A=-2$, $B=3$, $C=4$ in the markscheme!)*"},{"type":"text","order":4,"title":"Integrate with transformed limits","content":"Now we substitute our decomposed form back into the integral and integrate term by term:\n$$\\int_1^6 \\left( -2u^{-1} + 3u^{-2} + 4(u+2)^{-1} \\right) du$$\n\nIntegrating each term gives:\n$$\\left[ -2\\ln|u| - \\frac{3}{u} + 4\\ln|u + 2| \\right]_1^6$$"},{"type":"text","order":5,"title":"Evaluate the definite integral","content":"Now, we apply the integration limits from $u = 1$ to $u = 6$:\n\n**Upper limit ($u = 6$):**\n$$-2\\ln 6 - \\frac{3}{6} + 4\\ln 8 = -2\\ln 6 - \\frac{1}{2} + 4\\ln 8$$\n\n**Lower limit ($u = 1$):**\n$$-2\\ln 1 - \\frac{3}{1} + 4\\ln 3 = -3 + 4\\ln 3$$\n\nSubtracting the lower limit term from the upper limit term:\n$$\\left( -\\frac{1}{2} - 2\\ln 6 + 4\\ln 8 \\right) - (-3 + 4\\ln 3)$$\n$$\\left( -\\frac{1}{2} - 2\\ln 6 + 4\\ln 8 \\right) - (-3 + 4\\ln 3) = \\frac{5}{2} + 4\\ln 8 - 2\\ln 6 - 4\\ln 3$$"},{"type":"text","order":6,"title":"Simplify using logarithm laws","content":"Let's simplify the logarithmic expressions to find the final exact value:\n\n- Write $8$ as $2^3$:\n$$4\\ln 8 = 4\\ln(2^3) = 12\\ln 2$$\n- Write $6$ as $2 \\times 3$:\n$$2\\ln 6 = 2(\\ln 2 + \\ln 3) = 2\\ln 2 + 2\\ln 3$$\n\nSubstitute these back into our expression:\n$$\\frac{5}{2} + 12\\ln 2 - (2\\ln 2 + 2\\ln 3) - 4\\ln 3 = \\frac{5}{2} + 12\\ln 2 - 2\\ln 2 - 2\\ln 3 - 4\\ln 3$$\n$$\\frac{5}{2} + 12\\ln 2 - 2\\ln 2 - 2\\ln 3 - 4\\ln 3 = \\frac{5}{2} + 10\\ln 2 - 6\\ln 3$$\n\nThus, the exact value of the integral is:\n$$\\frac{5}{2} + 10\\ln 2 - 6\\ln 3$$"}]}],"generatedAt":"2026-06-09T03:16:33.461Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null}],"subjectCode":"a_level_maths__wjec","summaries":[{"id":"4792f464-c2f0-431d-8492-3b7e99a1453c","difficultyBand":"medium","calculatorUse":null,"totalMarks":12,"questionType":"LA","isCalculation":true},{"id":"5efaf326-4e1e-4803-b747-4f43fd8c194d","difficultyBand":"medium","calculatorUse":null,"totalMarks":8,"questionType":"LA","isCalculation":true},{"id":"6fa43b80-9cf1-433b-8a0e-ad3def83376a","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"90ed5e59-5f28-4250-b4b7-03ec6853d1c3","difficultyBand":"medium","calculatorUse":null,"totalMarks":10,"questionType":"LA","isCalculation":true},{"id":"22a686d1-c267-4b75-8ec2-56e5dc2f29bc","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"4a1af557-b6eb-4553-b064-f108c15794cb","difficultyBand":"medium","calculatorUse":null,"totalMarks":10,"questionType":"LA","isCalculation":true},{"id":"4cc4bca5-0208-440f-88f1-8d7390f3c68c","difficultyBand":"medium","calculatorUse":null,"totalMarks":8,"questionType":"LA","isCalculation":true},{"id":"8e7027a7-8b8e-4202-b7e8-8ccfef4b127f","difficultyBand":"medium","calculatorUse":null,"totalMarks":10,"questionType":"LA","isCalculation":true},{"id":"b110c2e2-8186-4ce9-bfa6-ae60cd7e7f6f","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"e9e83363-2d39-4903-8bb7-db90859b3c62","difficultyBand":"hard","calculatorUse":null,"totalMarks":8,"questionType":"LA","isCalculation":true},{"id":"38df22e6-68a0-4d39-bfb9-43461d32391c","difficultyBand":"medium","calculatorUse":null,"totalMarks":5,"questionType":"LA","isCalculation":true},{"id":"0cba29e6-3d13-4501-a510-cb4d1b065743","difficultyBand":"hard","calculatorUse":null,"totalMarks":8,"questionType":"LA","isCalculation":true},{"id":"0833676d-06e1-4cee-8f05-96bc3f2c4c28","difficultyBand":"medium","calculatorUse":null,"totalMarks":9,"questionType":"LA","isCalculation":true},{"id":"53df1b90-ed6c-48af-8019-102c532f0fec","difficultyBand":"medium","calculatorUse":null,"totalMarks":10,"questionType":"LA","isCalculation":true},{"id":"62aef9ac-23a6-4a18-ac18-832125e6c346","difficultyBand":"medium","calculatorUse":null,"totalMarks":9,"questionType":"LA","isCalculation":true},{"id":"6d8b07fe-9080-47ea-b452-aa30060e53e3","difficultyBand":"medium","calculatorUse":null,"totalMarks":10,"questionType":"LA","isCalculation":true},{"id":"7202e097-9bb4-4b42-b361-f64d00a025e0","difficultyBand":"medium","calculatorUse":null,"totalMarks":9,"questionType":"LA","isCalculation":true},{"id":"a918cfd1-d312-4489-9b8c-412af2144780","difficultyBand":"medium","calculatorUse":null,"totalMarks":10,"questionType":"LA","isCalculation":true},{"id":"ea5ab72f-05a3-4992-8dd9-fd7a66fab4f1","difficultyBand":"medium","calculatorUse":null,"totalMarks":9,"questionType":"LA","isCalculation":true}],"title":"3.7.5 Integration (A-level only)","topicId":"5d812869-4b4f-449f-89fd-18fec73d10fa","topicSidebar":"$L3a","topicSlug":"integration-10","topicTitle":"3.7.5 Integration (A-level only)"}]}],"$L3b","$L3c"]}]

3.7.5 Integration (A-level only)

Question 2

3.7.5 Integration (A-level only) Questions

3.7.5 Integration (A-level only)

Question 2

3.7.5 Integration (A-level only) Questions