Integration Involving Exponentials and Logarithms

A LevelAQAEdexcelOCR

Integration Involving Exponentials and Logarithms

Previously we have seen that the function $e^{x}$ is its own derivative. This also means that it is its own integral.

We have also previously seen that our rule for $x^{n}$ does not work for $n=-1$ . On this page we shall discover the integral of $x^{-1}$ .

Make sure you are happy with the following topics before continuing.

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Integrating the Exponential Function

The derivative of $e^{x}$ is $e^{x}$ . It follows that:

$int e^{x}dx=e^{x}+c$

Don’t forget $+c$ .

Indeed, more generally:

$int e^{ax+b}dx=dfrac{1}{a}e^{ax+b}+c$

Example: The integral of $e^{3x+4}$ is $dfrac{1}{3}e^{3x+4} + c$

Integrating to the Natural Logarithm

Generally, $int x^{n}dx=dfrac{1}{n+1}x^{n+1}+c$ . However, when $n=-1$ , this involves dividing by $0$ . So another function to integrate to must be found. This is the natural logarithm.

${LARGE int}dfrac{1}{x}dx=ln(|x|)+c$

This rule also comes in two other forms:

${LARGE int}dfrac{1}{ax+b}dx=dfrac{1}{a}ln(|ax+b|)+c$

${LARGE int}dfrac{f'(x)}{f(x)}dx=ln(|f(x)|)+c$

Examples: ${LARGE int}dfrac{1}{6-7x}dx=-dfrac{1}{7}ln(|6-7x|)+c$

${LARGE int}dfrac{3cos(3x)}{sin(3x)}dx=ln(|sin(3x)|)+c$ because the derivative of $sin(3x)$ is $3cos(3x)$ so it is of the form ${LARGE int}dfrac{f'(x)}{f(x)}dx$

Integrating Partial Fractions

To integrate partial fractions, split them into multiple terms then apply the appropriate integration rule to each term.

Example: Integrate $dfrac{5x+4}{(2x+1)(x+2)}$

Step 1: split it into partial fractions.

$dfrac{5x+4}{(2x+1)(x+2)}=dfrac{A}{2x+1}+dfrac{B}{x+2}$

$5x+4=A(x+2)+B(2x+1)$

$5x+4=Ax+2A+2Bx+B$

$A+2B=5$

$2A+B=4$

$2A+4B=10$

$3B=6$

$B=2$

$2A+2=4$

$2A=2$

$A=1$

$dfrac{5x+4}{(2x+1)(x+2)}=dfrac{1}{2x+1}+dfrac{2}{x+2}$

Step 2: Put it into the integral.

$begin{aligned}intdfrac{5x+4}{(2x+1)(x+2)}dx&=intdfrac{1}{2x+1}+dfrac{2}{x+2}dx[1.2em]&=intdfrac{1}{2x+1}dx+2intdfrac{1}{x+2}dxend{aligned}$

Step 3: Apply the rule to each term:

${LARGE int}dfrac{1}{2x+1}dx=dfrac{1}{2}ln(|2x+1|)+c$

${LARGE int}dfrac{1}{x+2}dx=ln(|x+2|)+c$

Step 4: Put it all together.

${LARGE int}dfrac{5x+4}{(2x+1)(x+2)}dx=dfrac{1}{2}ln(|2x+1|)+2ln(|x+2|)+c$

Integration Involving Exponentials and Logarithms Example Questions

Question 1: Integrate:

i) $e^{2x}$

ii) $3e^{5x+1}$

iii) $12e^{9-28x}$

[3 marks]

$begin{aligned}int e^{2x}dx=dfrac{1}{2}e^{2x}+cend{aligned}$

ii)

$begin{aligned}int left( 3e^{5x+1}right) dx&=left( 3timesdfrac{1}{5}e^{5x+1}right) +c[1.2em]&=dfrac{3}{5}e^{5x+1}+cend{aligned}$

iii)

$begin{aligned}int12e^{9-28x}dx&=left( 12times -dfrac{-}{28}e^{9-28x}right) +c[1.2em]&=-dfrac{12}{28}e^{9-28x}+c[1.2em]&=-dfrac{3}{7}e^{9-28x}+cend{aligned}$

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Question 2: Integrate:

i) $dfrac{1}{3x+1}$

ii) $dfrac{3}{x+4}$

iii) $dfrac{-2}{6-11x}$

[3 marks]

$begin{aligned}intdfrac{1}{3x+1}dx=dfrac{1}{3}ln(|3x+1|)+cend{aligned}$

ii)

$begin{aligned}intdfrac{3}{x+4}dx&=left( 3timesintdfrac{1}{x+4}right) +c[1.2em]&=3ln(|x+4|)+cend{aligned}$

iii)

$begin{aligned}intdfrac{-2}{6-11x}dx&=intdfrac{2}{11x-6}dx[1.2em]&=2timesintdfrac{1}{11x-6}dx[1.2em]&=left( 2timesdfrac{1}{11}ln(|11x-6|)right) +c[1.2em]&=dfrac{2}{11}ln(|11x-6|)+cend{aligned}$

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Question 3: Integrate:

i) $dfrac{6x+5}{3x^{2}+5x+4}$

ii) $dfrac{x^{3}}{x^{4}+5}$

iii) $dfrac{cos(2x)}{sin(x)cos(x)}$

[6 marks]

$begin{aligned}intdfrac{6x+5}{3x^{2}+5x+4}dx&=intdfrac{dfrac{d}{dx}(3x^{2}+5x+4)}{3x^{2}+5x+4}dx[1.2em]&=ln(|3x^{2}+5x+4|)+cend{aligned}$

ii)

$begin{aligned}intdfrac{x^{3}}{x^{4}+5}dx&=dfrac{1}{4}intdfrac{4x^{3}}{x^{4}+5}dx[1.2em]&=dfrac{1}{4}intdfrac{dfrac{d}{dx}(x^{4}+5)}{x^{4}+5}dx[1.2em]&=dfrac{1}{4}ln(|x^{4}+5|)+cend{aligned}$

iii)

$begin{aligned}&intdfrac{cos(2x)}{sin(x)cos(x)}dx=intdfrac{cos^{2}(x)-sin^{2}(x)}{sin(x)cos(x)}dx[1.2em]&=intdfrac{cos(x)dfrac{d}{dx}(sin(x))+sin(x)dfrac{d}{dx}(cos(x))}{sin(x)cos(x)}dx[1.2em]&=intdfrac{dfrac{d}{dx}(sin(x)cos(x))}{sin(x)cos(x)}dx[1.2em]&=ln(|sin(x)cos(x)|)+cend{aligned}$

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Question 4: Integrate $dfrac{11x+7}{3x^{2}+7x+2}$

[4 marks]

Split into partial fractions:

$begin{aligned}dfrac{11x+7}{3x^{2}+7x+2}&=dfrac{11x+7}{(3x+1)(x+2)}[1.2em]&=dfrac{A}{3x+1}+dfrac{B}{x+2}end{aligned}$

$11x+7=A(x+2)+B(3x+1)$

$11x+7=Ax+2A+3Bx+B$

$A+3B=11$

$2A+B=7$

$2A+6B=22$

$5B=15$

$B=3$

$2A+3=7$

$2A=4$

$A=2$

$dfrac{11x+7}{3x^{2}+7x+2}=dfrac{2}{3x+1}+dfrac{3}{x+2}$

Now integrate:

$begin{aligned}intdfrac{11x+7}{3x^{2}+7x+2}dx&=intleft( dfrac{2}{3x+1}+dfrac{3}{x+2}right) dx[1.2em]&=dfrac{2}{3}ln(|3x+1|)+3ln(|x+2|)+cend{aligned}$

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Additional Resources

Exam Tips Cheat Sheet

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Formula Booklet

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Specification Points Covered

H2 – Integrate $e^{kx}$ , $dfrac{1}{x}$ , $sin(kx)$ , $cos(kx)$ and related sums, differences and constant multiples
H6 – Integrate using partial fractions that are linear in the denominator