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Integration by substitution and by parts questions

The chain rule and the product rule each run backwards into a method. Substitution relabels an integral until it looks like something the shelf recognises, and parts trades one integral for a hopefully easier one, with the logarithm's own integral as its most famous conquest.

8 original questions · 32 marks · the integration by substitution and by parts notes · Integration

Every question here is written for this library rather than taken from a past paper. Write your answer out before opening the worked one: the answers award marks point by point, and the marks are easier to see when you have something of your own to compare against.

• 2 marks · find · warm-up

  Find ∫2x(x² + 1)³ dx by recognising a reverse chain pattern.

  Worked answer

  The bracket's derivative, 2x, is sitting alongside as a factor, so the integrand is exactly what the chain rule leaves behind. The integral is (x² + 1)⁴/4 + c, the next power up divided by that power. M1 for the recognition, A1 for the answer including the constant.

• 2 marks · find · warm-up

  Find ∫2x/(x² + 3) dx.

  Worked answer

  The numerator is exactly the derivative of the denominator, so the f′/f pattern applies and the integral is ln(x² + 3) + c. M1 A1. No modulus signs are needed, since x² + 3 is positive for every real x.

• 5 marks · use

  Use the substitution u = x² + 1 to find the exact value of ∫₀² x(x² + 1)³ dx.

  Worked answer

  From u = x² + 1, du = 2x dx, so x dx = du/2. The limits convert as well: x = 0 gives u = 1 and x = 2 gives u = 5. The integral becomes ½∫₁⁵ u³ du = ½[u⁴/4] from 1 to 5 = (625 − 1)/8 = 78. M1 for du/dx, M1 for the substitution, A1 for the integral fully in u with converted limits, M1 for integrating, A1 for 78. Carrying the old limits 0 and 2 into a u-integral gives (16 − 1)/8 and is the commonest way to lose the last three marks.

• 4 marks · use

  Use integration by parts to find ∫xe^2x dx.

  Worked answer

  Choose the factor that simplifies when differentiated, so u = x and dv/dx = e^2x. Then ∫xe^2x dx = (x/2)e^2x − ∫(1/2)e^2x dx = (x/2)e^2x − e^2x/4 + c. M1 for a correct application of the formula, A1 for the first stage, M1 for integrating the remaining term, A1 for the answer with its constant. The trade turned a product into a bare exponential, and choosing which factor plays which role is the whole judgement of the method.

• 4 marks · use

  Use integration by parts to find ∫x ln x dx.

  Worked answer

  Here the logarithm has to be the differentiated factor, because ∫ln x dx is not a standard result you can quote. Take u = ln x and dv/dx = x. Then ∫x ln x dx = (x²/2) ln x − ∫(x²/2)(1/x) dx = (x²/2) ln x − x²/4 + c. M1 for the choice and the formula, A1 for the first stage, M1 for simplifying x²/2 × 1/x to x/2, A1 for the answer. Whenever a logarithm turns up in a parts question it takes the u role, without exception at this level.

• 4 marks · find

  Find the exact value of ∫₀^π/2 x sin x dx.

  Worked answer

  Parts with u = x and dv/dx = sin x gives [−x cos x]₀^π/2 + ∫₀^π/2 cos x dx = [−x cos x + sin x] from 0 to π/2. At the upper limit that is −(π/2)(0) + 1 = 1, and at the lower limit it is 0 + 0 = 0, so the value is exactly 1. M1 for the parts, A1 for −x cos x + sin x, M1 for substituting the limits, A1 for 1. The boundary term vanishes at both ends for different reasons, cos (π/2) = 0 at the top and x = 0 at the bottom, and losing the sign on −∫(−cos x) dx is the usual slip.

• 6 marks · use · stretch

  Use the substitution u = 1 + √x to show that ∫₁⁴ 1/(1 + √x) dx = 2 − 2 ln (3/2).

  Worked answer

  From u = 1 + √x, √x = u − 1 and so x = (u − 1)², giving dx = 2(u − 1) du. The limits move too: x = 1 gives u = 2 and x = 4 gives u = 3. The integral becomes ∫₂³ 2(u − 1)/u du = 2∫₂³ (1 − 1/u) du = 2[u − ln u] from 2 to 3 = 2[(3 − ln 3) − (2 − ln 2)] = 2[1 − ln (3/2)] = 2 − 2 ln (3/2), as required. M1 for x in terms of u, M1 for dx, A1 for the new limits, M1 for the integrand fully in u, M1 for integrating, A1 for the printed form. Splitting (u − 1)/u into 1 − 1/u is the step that makes it integrable. Left as a single fraction it goes nowhere.

• 5 marks · find · stretch

  Find the exact value of ∫₀¹ x² e^x dx, giving your answer in terms of e.

  Worked answer

  Parts with u = x² and dv/dx = e^x gives x²e^x − 2∫x e^x dx. The remaining integral needs parts a second time, giving ∫x e^x dx = x e^x − e^x. So the antiderivative is x²e^x − 2x e^x + 2e^x. At x = 1 that is e − 2e + 2e = e, and at x = 0 it is 0 − 0 + 2 = 2, so the value is e − 2. M1 for the first parts, A1 for the first stage, M1 for the second parts, A1 for the full antiderivative, A1 for e − 2. Each application drops the power on x by one, so a polynomial of degree n needs n rounds and the sign alternates as you go.

The same practice on paper: the printable workbook for this topic, questions and a worked answer book.

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