# Evaluate the following integrals:

Question:

Evaluate the following integrals:

$\int \frac{e^{x}}{e^{2 x}+5 e^{x}+6} d x$

Solution:

let $I=\int \frac{e^{x}}{e^{2 x}+5 e^{x}+6} d x$

Let $\mathrm{e}^{\mathrm{x}}=\mathrm{t} \ldots \ldots$ (i)

$\Rightarrow \mathrm{e}^{\mathrm{x}} \mathrm{dx}=\mathrm{dt}$

$=\int \frac{1}{t^{2}+5 t+6} d t$

$=\int \frac{1}{t^{2}+2 t \times \frac{5}{2}+\left(\frac{5}{2}\right)^{2}-\left(\frac{5}{2}\right)^{2}+6} d t$

$=\int \frac{1}{\left(t+\frac{5}{2}\right)^{2}-\frac{1}{4}} d t$

Let $\mathrm{t}+\frac{5}{2}=\mathrm{u}$

$\Rightarrow \mathrm{dt}=\mathrm{du}$

$\mathrm{SO}$,

$I=\int \frac{1}{u^{2}-\left(\frac{1}{2}\right)^{2}} d u$

$I=\frac{1}{2 \times \frac{1}{2}} \log \left|\frac{u-\frac{1}{2}}{u+\frac{1}{2}}\right|+c$

[since, $\left.\int \frac{1}{x^{2}-(a)^{2}} d x=\frac{1}{2 \times a} \log \left|\frac{x-a}{x+a}\right|+c\right]$

$I=\log \left|\frac{2 u-1}{2 u+1}\right|+c$

$I=\log \left|\frac{2\left(t+\frac{5}{2}\right)-1}{2\left(t+\frac{5}{2}\right)+1}\right|+c[$ using $(i)]$

$I=\log \left|\frac{e^{x}+2}{e^{x}+3}\right|+c[$ using $(i i)]$