Learn7357Paper 3 · Pure and StatisticsPure content (sections A–J)Exact definite integrals and logarithms

Pure content (sections A–J) · Subtopic 49 of 66

Exact Definite Integrals and Logarithms

Evaluating definite integrals exactly when the answer involves $e$ , $\ln$ or $\pi$ ; simplifying logarithmic answers; spotting $\dfrac{f'(x)}{f(x)}$ ; and integrating with partial fractions.

Cambridge 9709 Paper 2Cambridge 9709 Paper 3Edexcel YMA01 WMA13Edexcel 9MA0 Paper 1Edexcel 9MA0 Paper 2AQA 7357 Paper 1≈ 45 minRevision 5 · updated 2026-10-27

Section 1 of 14

Learning objectives

Evaluate definite integrals of exponential, reciprocal and trigonometric functions exactly.
Simplify answers with the log laws, for example $\ln9-\ln3=\ln3$ .
(Paper 3) Recognise $\displaystyle\int\frac{kf'(x)}{f(x)}\,dx=k\ln|f(x)|+c$ , and integrate using partial fractions.
Find an unknown limit from the value of a definite integral.

Pause and think

Evaluate $\displaystyle\int_1^3\frac1x\,dx$ exactly. Why is " $1.0986$ " not the answer an exam wants?

Section 2 of 14

Core idea

Start hereAnswers like ln 16

With $e$ and $\ln$ in play, definite integrals give exact answers like $\ln16$ or $e^2-1$ . A calculator decimal hides the structure; an exact answer shows it. Log laws do the tidying.

The big idea

Evaluate $[F(x)]_a^b$ as usual, then simplify with $\ln a-\ln b=\ln\frac ab$ , $k\ln a=\ln a^k$ and $e^{\ln a}=a$ . Leave answers exact unless told otherwise.

1. Log results

Area under y = 2/x from 1 to 4 equals ln 16

An exact answer like

\ln16

is preferred unless a decimal is asked for.

Worked example $\int_1^4\frac2x\,dx$

Evaluate exactly.

Goal: Single log.
Strategy: Integrate; log laws.

$[2\ln x]_1^4=2\ln4-2\ln1$
$\ln1=0$ .
$=\ln4^2=\ln16$
Power law.

Answer

\ln16

✓ Check

\ln16=2.77

; area under

\frac2x

over width

3

with heights

2

down to

0.5

: plausible ✓.

Worked example $\int_2^5\frac3{3x-1}dx$

Evaluate.

Goal: Single log.
Strategy: $\frac1a\ln|ax+b|$ with $a=3$ .

$[\ln(3x-1)]_2^5=\ln14-\ln5$
$3\cdot\frac13=1$ .
$=\ln\tfrac{14}5$
Quotient law.

Answer

\ln\frac{14}5

✓ Check

\approx1.03

; integrand between

\frac3{14}

and

\frac35

over width

3

: between

0.64

and

1.8

✓.

Worked example $\int_0^{\ln3}e^{2x}dx$

Evaluate.

Goal: Exact number.
Strategy: $e^{2\ln3}=e^{\ln9}=9$ .

$\left[\tfrac12e^{2x}\right]_0^{\ln3}=\tfrac12\left(e^{2\ln3}-1\right)$
Integrate.
$=\tfrac12(9-1)=4$
Simplify.

Answer

4

✓ CheckNumerically

\int_0^{1.0986}e^{2x}dx\approx4.000

✓.

Worked exampleWith partial fractions

Evaluate $\int_2^3\frac5{(x-1)(x+4)}dx$ .

Goal: Single log.
Strategy: Split, integrate, combine.

$\frac5{(x-1)(x+4)}=\frac1{x-1}-\frac1{x+4}$
Cover-up.
$[\ln(x-1)-\ln(x+4)]_2^3=(\ln2-\ln7)-(\ln1-\ln6)$
Integrate; substitute.
$=\ln\frac{2\times6}7=\ln\frac{12}7$
Combine.

Answer

\ln\frac{12}7

✓ Check

\approx0.539

; positive since the integrand is positive on

[2,3]

✓.

Common misconception $\ln a-\ln b\neq\ln(a-b)$ It's

\ln\frac ab

Pause and think

$\int_0^{\pi/6}\cos3x\,dx$ ?

Summary

Combine logs at the end.
$e^{k\ln a}=a^k$ .

Exact value: Not rounded; with ln, e, π or surds.

HistoryMercator’s series. In 1668 Nicholas Mercator found the area under

y=\dfrac{1}{1+x}

by expanding it as a series, discovering

\ln(1+x)=x-\tfrac{x^2}{2}+\tfrac{x^3}{3}-\dots

. This made it possible to compute logarithms far more easily than with Napier's original method.

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