IGCSE · Practice questions · Integration

Add one, divide, plus c.

Six original Cambridge-style questions on indefinite integration: powers including \(\dfrac{1}{x}\), rewriting before integrating, \((ax + b)^{n}\), \(\dfrac{1}{ax + b}\), \(\mathrm{e}^{ax + b}\) and trigonometric functions, finding a curve from its gradient and a 'hence' integral, all checked by differentiating.

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

The method

Rewrite, integrate, divide by \(a\), add \(c\).

Rewrite as a sum of terms: roots as fractional powers, \(\dfrac{1}{x^{n}}\) as \(x^{-n}\), brackets expanded, fractions split over a single term.
Powers: add one to the power and divide by the new power; \(\dfrac{1}{x}\) gives \(\ln|x|\).
Linear inside \(ax + b\): integrate as if it were \(x\), then divide by \(a\). \(\sin\) gives \(-\cos\), \(\cos\) gives \(\sin\), \(\sec^{2}\) gives \(\tan\).
Add \(c\) to every indefinite integral; find it from a point by substituting into \(y\).
Hence: read the given derivative backwards, divide by any constant factor and integrate the extra terms. Check by differentiating.

Recall

[2 marks]

Find \(\displaystyle\int \left(5x^{4} - 6x + \frac{3}{x}\right)\mathrm{d}x\).

Term by term: \(\dfrac{5x^{5}}{5} - \dfrac{6x^{2}}{2} + 3\ln|x|\). ✓
\(x^{5} - 3x^{2} + 3\ln|x| + c\). ✓

Calculation

[3 marks]

Find \(\displaystyle\int \frac{(\sqrt{x} - 2)^{2}}{\sqrt{x}}\,\mathrm{d}x\).

Expand: \((\sqrt{x} - 2)^{2} = x - 4\sqrt{x} + 4\). ✓
Divide each term by \(x^{\frac{1}{2}}\): \(x^{\frac{1}{2}} - 4 + 4x^{-\frac{1}{2}}\). ✓
Integrate: \(\dfrac{x^{\frac{3}{2}}}{\frac{3}{2}} - 4x + \dfrac{4x^{\frac{1}{2}}}{\frac{1}{2}} + c = \dfrac{2}{3}x^{\frac{3}{2}} - 4x + 8x^{\frac{1}{2}} + c\). ✓

Calculation

[4 marks]

Find

(a) \(\displaystyle\int 10(2x - 3)^{4}\,\mathrm{d}x\),

(b) \(\displaystyle\int \frac{4}{1 - 2x}\,\mathrm{d}x\),

(a) \(\dfrac{10(2x - 3)^{5}}{2 \times 5} + c = (2x - 3)^{5} + c\). ✓
(b) Here \(a = -2\): \(4 \times \dfrac{1}{-2}\ln|1 - 2x| + c = -2\ln|1 - 2x| + c\). ✓
(c) Here \(a = -3\): \(-\dfrac{1}{3}\mathrm{e}^{1 - 3x} + c\). ✓
Check (b): \(\dfrac{\mathrm{d}}{\mathrm{d}x}\left(-2\ln|1 - 2x|\right) = -2 \times \dfrac{-2}{1 - 2x} = \dfrac{4}{1 - 2x}\). ✓

Calculation

[4 marks]

Find \[\displaystyle\int \left(6\cos 3x - 2\sin\frac{x}{2} + \sec^{2}(2x + 1)\right)\mathrm{d}x\].

\(6\cos 3x\) integrates to \(6 \times \dfrac{1}{3}\sin 3x = 2\sin 3x\). ✓
\(-2\sin\dfrac{x}{2}\) has \(a = \dfrac{1}{2}\): \(-2 \times \left(-\dfrac{\cos\frac{x}{2}}{\frac{1}{2}}\right) = 4\cos\dfrac{x}{2}\). ✓
\(\sec^{2}(2x + 1)\) integrates to \(\dfrac{1}{2}\tan(2x + 1)\). ✓
Total: \(2\sin 3x + 4\cos\dfrac{x}{2} + \dfrac{1}{2}\tan(2x + 1) + c\). ✓

Analysis

[5 marks]

A curve has \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 3\sqrt{x} - \dfrac{6}{x^{2}}\) for \(x > 0\), and passes through the point \((1, 3)\).

(a) Find the equation of the curve.

(b) Find the \(y\)-coordinate of the point on the curve where \(x = 4\).

(a) Write \(3x^{\frac{1}{2}} - 6x^{-2}\) and integrate: \(y = \dfrac{3x^{\frac{3}{2}}}{\frac{3}{2}} - \dfrac{6x^{-1}}{-1} + c = 2x^{\frac{3}{2}} + \dfrac{6}{x} + c\). ✓
At \((1, 3)\): \(3 = 2 + 6 + c\), so \(c = -5\). ✓
\(y = 2x^{\frac{3}{2}} + \dfrac{6}{x} - 5\). ✓
(b) \(x = 4\): \(4^{\frac{3}{2}} = 8\), so \(y = 16 + \dfrac{6}{4} - 5 = 12.5\). ✓

Analysis

[6 marks]

(a) Show that \(\dfrac{\mathrm{d}}{\mathrm{d}x}(x^{2}\ln x) = 2x\ln x + x\).

(b) Hence find \(\displaystyle\int x\ln x\,\mathrm{d}x\).

(a) Product rule: \(2x \times \ln x + x^{2} \times \dfrac{1}{x} = 2x\ln x + x\). ✓
(b) Reading (a) backwards: \(\displaystyle\int (2x\ln x + x)\,\mathrm{d}x = x^{2}\ln x + c\). ✓
So \(\displaystyle 2\int x\ln x\,\mathrm{d}x = x^{2}\ln x - \int x\,\mathrm{d}x = x^{2}\ln x - \frac{x^{2}}{2} + c\). ✓
Divide by 2: \(\displaystyle\int x\ln x\,\mathrm{d}x = \frac{1}{2}x^{2}\ln x - \frac{1}{4}x^{2} + c\). ✓
(c) \[\dfrac{\mathrm{d}}{\mathrm{d}x}\left(\dfrac{1}{2}x^{2}\ln x - \dfrac{1}{4}x^{2}\right) = x\ln x + \dfrac{1}{2}x - \dfrac{1}{2}x = x\ln x\], as required. ✓

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.