# 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.

A LevelAQAEdexcelOCR

## 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 LevelAQAEdexcelOCR

@mmerevise

## 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 LevelAQAEdexcelOCR

## 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 LevelAQAEdexcelOCR

## Integration Involving Exponentials and Logarithms 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}

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