c{8}{3}(x+2)(2x+1)$$\n\nThis is in the required form $y = f(x)$."}]}],"generatedAt":"2026-06-09T05:50:53.589Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null},{"id":"5efaf326-4e1e-4803-b747-4f43fd8c194d","revisionId":"4ccdbfcf-d218-458f-841a-c7950948bb30","questionNumber":null,"questionType":"LA","specification":"","difficulty":3,"difficultyBand":"medium","totalMarks":8,"sourceTitle":null,"calculatorUse":null,"parts":[{"id":"5344b1d7-8e93-4ea5-bbad-8bb294ce7b32","label":"a","content":"Find the exact value of $\\int_0^1 \\frac{4x^2 + 13x + 12}{(x+2)^2(x+1)} dx$","markscheme":"- Sets up partial fractions: $\\frac{A}{x+2} + \\frac{B}{(x+2)^2} + \\frac{C}{x+1}$ M1\n- Finds constants: $A=1, B=-2, C=3$ A1 A1\n- Integrates: $\\left[ \\ln|x+2| + 2(x+2)^{-1} + 3\\ln|x+1| \\right]_0^1$ M1 A1\n- Substitutes limits: $(\\ln 3 + \\frac{2}{3} + 3\\ln 2) - (\\ln 2 + 1)$ M1\n- Simplifies: $\\ln 3 + 2\\ln 2 - \\frac{1}{3}$ M1\n- Final exact value: $\\ln 12 - \\frac{1}{3}$ 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}{x+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=3$","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","code":"A"},{"id":"p6","marks":1,"description":"Substitutes limits 0 and 1 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 12 - \\frac{1}{3}","\\ln 12 - 0.333...","\\ln 3 + 2\\ln 2 - \\frac{1}{3}"],"generatedAt":"2026-05-06T14:04:42.809Z","modelVersion":"gemini-3-flash-preview"},"explanation":{"methods":[{"id":"direct_partial_fractions","label":"Direct Partial Fractions","steps":[{"type":"text","order":0,"title":"Set up partial fractions","content":"To integrate this rational function, we first decompose the integrand into partial fractions. Since the denominator has a repeated linear factor $(x+2)^2$ and a distinct linear factor $(x+1)$, we write it in the form:\n\n$$\\frac{4x^2 + 13x + 12}{(x+2)^2(x+1)} = \\frac{A}{x+2} + \\frac{B}{(x+2)^2} + \\frac{C}{x+1}$$\n\nThis is a standard template for algebraic fractions with repeated factors."},{"type":"text","order":1,"title":"Form the basic equation","content":"To find the constants $A$, $B$, and $C$, we multiply both sides of the identity by the common denominator $(x+2)^2(x+1)$:\n\n$$4x^2 + 13x + 12 = A(x+2)(x+1) + B(x+1) + C(x+2)^2$$"},{"type":"text","order":2,"title":"Solve for the constants","content":"We can choose convenient values of $x$ to isolate and solve for each constant.\n\nFirst, let $x = -2$ to eliminate the terms with $(x+2)$:\n$$\\begin{aligned} 4(-2)^2 + 13(-2) + 12 &= B(-2+1) \\\\ 16 - 26 + 12 &= -B \\\\ 2 &= -B \\implies B = -2 \\end{aligned}$$\n\nNext, let $x = -1$ to eliminate the terms with $(x+1)$:\n$$\\begin{aligned} 4(-1)^2 + 13(-1) + 12 &= C(-1+2)^2 \\\\ 4 - 13 + 12 &= C(1)^2 \\\\ 3 &= C \\implies C = 3 \\end{aligned}$$\n\nFinally, compare the coefficients of $x^2$ on both sides to find $A$:\n$$\\begin{aligned} 4 &= A + C \\\\ 4 &= A + 3 \\implies A = 1 \\end{aligned}$$\n\nSo, our partial fractions are:\n$$\\frac{1}{x+2} - \\frac{2}{(x+2)^2} + \\frac{3}{x+1}$$"},{"type":"text","order":3,"title":"Rewrite and integrate","content":"Now we substitute our decomposed fractions back into the integral:\n\n$$\\int_0^1 \\left( \\frac{1}{x+2} - \\frac{2}{(x+2)^2} + \\frac{3}{x+1} \\right) dx$$

3.6.5 Integration (A-level only)

Question 2

3.6.5 Integration (A-level only) Questions

3.6.5 Integration (A-level only)

Question 2

3.6.5 Integration (A-level only) Questions