# Integration Involving Exponentials and Logarithms

A LevelAQAEdexcelOCRAQA 2022Edexcel 2022OCR 2022

## 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.

A Level

## 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 $\dfrac{1}{3}e^{3x+4} + c$

A Level

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

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

A Level

## 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}\int\dfrac{5x+4}{(2x+1)(x+2)}dx&=\int\dfrac{1}{2x+1}+\dfrac{2}{x+2}dx\\[1.2em]&=\int\dfrac{1}{2x+1}dx+2\int\dfrac{1}{x+2}dx\end{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.

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

A Level

## Example Questions

i)

\begin{aligned}\int e^{2x}dx=\dfrac{1}{2}e^{2x}+c\end{aligned}

ii)

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

iii)

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

i)

\begin{aligned}\int\dfrac{1}{3x+1}dx=\dfrac{1}{3}\ln(|3x+1|)+c\end{aligned}

ii)

\begin{aligned}\int\dfrac{3}{x+4}dx&=\left( 3\times\int\dfrac{1}{x+4}\right) +c\\[1.2em]&=3\ln(|x+4|)+c\end{aligned}

iii)

\begin{aligned}\int\dfrac{-2}{6-11x}dx&=\int\dfrac{2}{11x-6}dx\\[1.2em]&=2\times\int\dfrac{1}{11x-6}dx\\[1.2em]&=\left( 2\times\dfrac{1}{11}\ln(|11x-6|)\right) +c\\[1.2em]&=\dfrac{2}{11}\ln(|11x-6|)+c\end{aligned}

i)

\begin{aligned}\int\dfrac{6x+5}{3x^{2}+5x+4}dx&=\int\dfrac{\dfrac{d}{dx}(3x^{2}+5x+4)}{3x^{2}+5x+4}dx\\[1.2em]&=\ln(|3x^{2}+5x+4|)+c\end{aligned}

ii)

\begin{aligned}\int\dfrac{x^{3}}{x^{4}+5}dx&=\dfrac{1}{4}\int\dfrac{4x^{3}}{x^{4}+5}dx\\[1.2em]&=\dfrac{1}{4}\int\dfrac{\dfrac{d}{dx}(x^{4}+5)}{x^{4}+5}dx\\[1.2em]&=\dfrac{1}{4}\ln(|x^{4}+5|)+c\end{aligned}

iii)

\begin{aligned}&\int\dfrac{\cos(2x)}{\sin(x)\cos(x)}dx=\int\dfrac{\cos^{2}(x)-\sin^{2}(x)}{\sin(x)\cos(x)}dx\\[1.2em]&=\int\dfrac{\cos(x)\dfrac{d}{dx}(\sin(x))+\sin(x)\dfrac{d}{dx}(\cos(x))}{\sin(x)\cos(x)}dx\\[1.2em]&=\int\dfrac{\dfrac{d}{dx}(\sin(x)\cos(x))}{\sin(x)\cos(x)}dx\\[1.2em]&=\ln(|\sin(x)\cos(x)|)+c\end{aligned}

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}\int\dfrac{11x+7}{3x^{2}+7x+2}dx&=\int\left( \dfrac{2}{3x+1}+\dfrac{3}{x+2}\right) dx\\[1.2em]&=\dfrac{2}{3}\ln(|3x+1|)+3\ln(|x+2|)+c\end{aligned}

A Level

A Level

A Level

A Level

A Level

## You May Also Like...

### A Level Maths Revision Cards

The best A level maths revision cards for AQA, Edexcel, OCR, MEI and WJEC. Maths Made Easy is here to help you prepare effectively for your A Level maths exams.

£14.99

### A Level Maths – Cards & Paper Bundle

A level maths revision cards and exam papers for Edexcel. Includes 2022 predicted papers based on the advance information released in February 2022! MME is here to help you study from home with our revision cards and practise papers.

From: £22.99

### Transition Maths Cards

The transition maths cards are a perfect way to cover the higher level topics from GCSE whilst being introduced to new A level maths topics to help you prepare for year 12. Your ideal guide to getting started with A level maths!

£8.99