AS & A Level · Practice questions · Integration by parts

Differentiate one, integrate the other.

Six original Cambridge-style questions on integration by parts, including logarithms, parts twice and inverse tan, with exact answers.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Choose u so the new integral is simpler.

Formula: \(\displaystyle\int u\dfrac{dv}{dx}\,dx = uv - \int v\dfrac{du}{dx}\,dx\).
u is \(\ln x\) or \(\tan^{-1}x\) if present, otherwise the power of \(x\).
Lone ln x: take \(\dfrac{dv}{dx} = 1\).
Twice for \(x^2\) times an exponential or trig function; bracket the second result.
Check by differentiating your answer.

Recall

[3 marks]

Find \(\displaystyle\int xe^{3x}\,dx\).

\(u = x\), \(v = \dfrac{1}{3}e^{3x}\). ✓
\(\dfrac{x}{3}e^{3x} - \displaystyle\int \dfrac{1}{3}e^{3x}\,dx = \dfrac{x}{3}e^{3x} - \dfrac{1}{9}e^{3x} + c\). ✓

Calculation

[4 marks]

Find the exact value of \(\displaystyle\int_0^{\pi} x\sin x\,dx\).

\(u = x\), \(v = -\cos x\): \(\displaystyle\int x\sin x\,dx = -x\cos x + \int \cos x\,dx = -x\cos x + \sin x\). ✓
\(\left[-x\cos x + \sin x\right]_0^{\pi} = (-\pi\cos\pi + 0) - 0 = \pi\). ✓

Calculation

[5 marks]

Show that \(\displaystyle\int_1^{2} x^2\ln x\,dx = \dfrac{8}{3}\ln 2 - \dfrac{7}{9}\).

\(u = \ln x\), \(\dfrac{du}{dx} = \dfrac{1}{x}\); \(\dfrac{dv}{dx} = x^2\), \(v = \dfrac{x^3}{3}\). ✓
\(\dfrac{x^3}{3}\ln x - \displaystyle\int \dfrac{x^3}{3}\cdot\dfrac{1}{x}\,dx = \dfrac{x^3}{3}\ln x - \dfrac{x^3}{9}\). ✓
\(\left(\dfrac{8}{3}\ln 2 - \dfrac{8}{9}\right) - \left(0 - \dfrac{1}{9}\right) = \dfrac{8}{3}\ln 2 - \dfrac{7}{9}\). ✓

Proof

[5 marks]

Show that \(\displaystyle\int_0^1 x^2e^{x}\,dx = e - 2\).

\(u = x^2\), \(v = e^{x}\): \(x^2e^{x} - \displaystyle\int 2xe^{x}\,dx\). ✓
\(\displaystyle\int 2xe^{x}\,dx = 2xe^{x} - 2e^{x}\). ✓
So \(\displaystyle\int x^2e^{x}\,dx = e^{x}(x^2 - 2x + 2)\). ✓
\(\left[e^{x}(x^2 - 2x + 2)\right]_0^1 = e(1) - 1(2) = e - 2\). ✓

Calculation

[5 marks]

Find the exact value of \(\displaystyle\int_0^1 \tan^{-1}x\,dx\).

\(u = \tan^{-1}x\), \(\dfrac{dv}{dx} = 1\), so \(v = x\). ✓
\(x\tan^{-1}x - \displaystyle\int \dfrac{x}{1+x^2}\,dx = x\tan^{-1}x - \dfrac{1}{2}\ln(1+x^2)\). ✓
\(\left[x\tan^{-1}x - \dfrac{1}{2}\ln(1+x^2)\right]_0^1 = \dfrac{\pi}{4} - \dfrac{1}{2}\ln 2\). ✓

Analysis

[8 marks]

The curve \(y = xe^{-x/2}\) has a maximum point \(M\).

(a) Find the exact coordinates of \(M\).

(b) Find the exact area of the region bounded by the curve, the \(x\)-axis and the line through \(M\) parallel to the \(y\)-axis, and give it to 3 significant figures.

(a) \(\dfrac{dy}{dx} = e^{-x/2} - \dfrac{x}{2}e^{-x/2} = e^{-x/2}\left(1 - \dfrac{x}{2}\right)\). ✓
\(\dfrac{dy}{dx} = 0 \Rightarrow x = 2\), \(y = 2e^{-1}\). So \(M = \left(2, \dfrac{2}{e}\right)\). ✓
(b) Area \(= \displaystyle\int_0^2 xe^{-x/2}\,dx\). Parts: \(u = x\), \(v = -2e^{-x/2}\). ✓
\(\displaystyle\int xe^{-x/2}\,dx = -2xe^{-x/2} + \int 2e^{-x/2}\,dx = -2xe^{-x/2} - 4e^{-x/2}\). ✓
\(\left[-2xe^{-x/2} - 4e^{-x/2}\right]_0^2 = \left(-\dfrac{4}{e} - \dfrac{4}{e}\right) - (-4) = 4 - \dfrac{8}{e}\). ✓
\(= 1.06\) square units (3 s.f.). ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.