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Learn9MA0Paper 2 · Pure Mathematics 2IntegrationIntegrating eˣ, 1/x and trigonometric functions
Integration · Subtopic 5 of 14

Integrating eˣ, 1/x and Trigonometric Functions

Reversing the new derivatives: ∫eax+b dx\int e^{ax+b}\,dx, ∫1ax+b dx\int\dfrac{1}{ax+b}\,dx (the missing case n=−1n=-1), and ∫sin⁡\int\sin, cos⁡\cos and sec⁡2\sec^2, always remembering to divide by aa.

Cambridge 9709 Paper 2Cambridge 9709 Paper 3Edexcel YMA01 WMA13Edexcel 9MA0 Paper 1Edexcel 9MA0 Paper 2AQA 7357 Paper 1≈ 40 minRevision 5 · updated 2026-10-27
Section 1 of 14

Learning objectives

  • Integrate eax+be^{ax+b}, 1ax+b\dfrac{1}{ax+b}, sin⁡(ax+b)\sin(ax+b), cos⁡(ax+b)\cos(ax+b) and sec⁡2(ax+b)\sec^2(ax+b).
  • Explain why ∫1x dx=ln⁡∣x∣+c\displaystyle\int\frac1x\,dx=\ln|x|+c.
  • Rewrite expressions (splitting fractions, expanding brackets) before integrating.
  • Check answers by differentiating.
Pause and think

In Pure 1 the rule ∫xn dx=xn+1n+1\displaystyle\int x^n\,dx=\frac{x^{n+1}}{n+1} failed for n=−1n=-1. Which function has derivative 1x\dfrac1x?

Section 2 of 14

Core idea

Start hereThe missing power

The power rule integrates every xnx^n... except x−1x^{-1}, which would give x00\frac{x^0}0. So what is ∫1x dx\int\frac1x\,dx? The area under y=1xy=\frac1x from 11 to bb turns out to be exactly ln⁡b\ln b. Reversing the new derivatives gives a whole family of standard integrals.

The big idea

∫ex dx=ex+c\int e^x\,dx=e^x+c; ∫1x dx=ln⁡∣x∣+c\int\frac1x\,dx=\ln|x|+c; ∫cos⁡x dx=sin⁡x+c\int\cos x\,dx=\sin x+c; ∫sin⁡x dx=−cos⁡x+c\int\sin x\,dx=-\cos x+c; ∫sec⁡2x dx=tan⁡x+c\int\sec^2x\,dx=\tan x+c. With ax+bax+b inside, divide by aa. ∫f′(x)f(x)dx=ln⁡∣f(x)∣+c\int\frac{f'(x)}{f(x)}dx=\ln|f(x)|+c.

1. The table

Area under y = 1/x from 1 to b is ln b
123456780.511.522.530xyy = 1/xarea = ln b
When b=e≈2.72b=e\approx2.72 the area is exactly 11. The area keeps growing as bb grows, but ever more slowly, just like ln⁡b\ln b.
∫eax+bdx=1aeax+b+c∫1ax+bdx=1aln⁡∣ax+b∣+c∫cos⁡(ax+b)dx=1asin⁡(ax+b)+c∫sin⁡(ax+b)dx=−1acos⁡(ax+b)+c∫sec⁡2(ax+b)dx=1atan⁡(ax+b)+c\begin{aligned}\int e^{ax+b}dx&=\tfrac1ae^{ax+b}+c&\int\frac1{ax+b}dx&=\tfrac1a\ln|ax+b|+c\\\int\cos(ax+b)dx&=\tfrac1a\sin(ax+b)+c&\int\sin(ax+b)dx&=-\tfrac1a\cos(ax+b)+c\\\int\sec^2(ax+b)dx&=\tfrac1a\tan(ax+b)+c\end{aligned}
Key takeawayWhy ∣x∣|x| For x<0x<0, ddxln⁡(−x)=−1−x=1x\frac{d}{dx}\ln(-x)=\frac{-1}{-x}=\frac1x too. So ln⁡∣x∣\ln|x| covers both signs.

2. Examples

Worked exampleThree standard integrals

Find ∫e3xdx\int e^{3x}dx, ∫42x−1dx\int\frac4{2x-1}dx, ∫sin⁡(2x+π3)dx\int\sin\left(2x+\frac\pi3\right)dx.

Goal
Antiderivatives.
Strategy
Table; divide by aa.
  1. 13e3x+c\tfrac13e^{3x}+c
    a=3a=3.
  2. 4⋅12ln⁡∣2x−1∣=2ln⁡∣2x−1∣+c4\cdot\tfrac12\ln|2x-1|=2\ln|2x-1|+c
    a=2a=2.
  3. −12cos⁡(2x+π3)+c-\tfrac12\cos\left(2x+\tfrac\pi3\right)+c
    a=2a=2; sine integrates to minus cosine.
AnswerAs shown.
✓ CheckDifferentiate 2ln⁡∣2x−1∣2\ln|2x-1|: 42x−1\frac{4}{2x-1} ✓.
Worked example∫xx2+4dx\int\frac{x}{x^2+4}dx by spotting f′/ff'/f

Integrate.

Goal
Antiderivative.
Strategy
The top is (nearly) the derivative of the bottom.
  1. ddx(x2+4)=2x;  x=12(2x)\frac{d}{dx}(x^2+4)=2x;\ \ x=\tfrac12(2x)
    Adjust the constant.
  2. 12∫2xx2+4dx=12ln⁡(x2+4)+c\tfrac12\int\frac{2x}{x^2+4}dx=\tfrac12\ln(x^2+4)+c
    ∫f′f=ln⁡∣f∣\int\frac{f'}f=\ln|f|; x2+4>0x^2+4>0.
Answer12ln⁡(x2+4)+c\frac12\ln(x^2+4)+c.
✓ CheckDifferentiate: 12⋅2xx2+4\frac12\cdot\frac{2x}{x^2+4} ✓.
Worked example∫tan⁡x dx\int\tan x\,dx

Integrate.

Goal
Antiderivative.
Strategy
Write as sin⁡xcos⁡x\frac{\sin x}{\cos x}; the top is minus the derivative of the bottom.
  1. ∫sin⁡xcos⁡xdx=−∫−sin⁡xcos⁡xdx\int\frac{\sin x}{\cos x}dx=-\int\frac{-\sin x}{\cos x}dx
    Introduce the derivative of cos⁡x\cos x.
  2. =−ln⁡∣cos⁡x∣+c=ln⁡∣sec⁡x∣+c=-\ln|\cos x|+c=\ln|\sec x|+c
    f′/ff'/f; log law.
Answerln⁡∣sec⁡x∣+c\ln|\sec x|+c.
✓ Checkddxln⁡sec⁡x=sec⁡xtan⁡xsec⁡x=tan⁡x\frac{d}{dx}\ln\sec x=\frac{\sec x\tan x}{\sec x}=\tan x ✓.
Common misconceptionSign of sine ∫sin⁡x dx=−cos⁡x\int\sin x\,dx=-\cos x (because ddxcos⁡x=−sin⁡x\frac{d}{dx}\cos x=-\sin x).
Pause and think

∫(e−x+1x)dx\int(e^{-x}+\frac1x)dx?

Summary

  • Standard table; divide by aa.
  • f′/f→ln⁡∣f∣f'/f\to\ln|f|.
Key words
Reverse chain rule
Recognising a derivative already present.
HistoryLogs of negative numbers. In 1712 Leibniz and Johann Bernoulli argued in letters about whether ln⁡(−1)\ln(-1) exists. Bernoulli said ln⁡(−x)=ln⁡x\ln(-x)=\ln x, since both have derivative 1x\tfrac1x. Euler finally sorted it out decades later. For integrating with real numbers, we write ln⁡∣x∣\ln|x|.
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