#### Explain solution RD Sharma class 12 chapter Indefinite Integrals exercise 18.16 question 4 maths

$log\left | \frac{e^{x}+2}{e^{x}+3} \right |+C$

Hint:

Use substitution method as well as special integration formula to solve this type of problem

Given:

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

Solution:

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

$Put\: \: e^{x}=t\Rightarrow e^{x}dx=dt$

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

\begin{aligned} &=\int \frac{1}{t^{2}+2 \cdot t \cdot \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{25}{4}+6} d t=\int \frac{1}{\left(t+\frac{5}{2}\right)^{2}-\frac{25+24}{4}} d t \end{aligned}

\begin{aligned} &=\int \frac{1}{\left(t+\frac{5}{2}\right)^{2}-\frac{1}{4}} d t \\ &=\int \frac{1}{\left(t+\frac{5}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}} d t \end{aligned}

$Put\: \: t+\frac{5}{2}=u\Rightarrow dt=du$

\begin{aligned} =\int \frac{1}{u^{2}-\left(\frac{1}{2}\right)^{2}} d u \end{aligned}

\begin{aligned} &=\frac{1}{2 \times \frac{1}{2}} \log \left|\frac{u-\frac{1}{2}}{u+\frac{1}{2}}\right|+C \quad\left[\because \int \frac{1}{x^{2}-a^{2}} d x=\frac{1}{2 a} \log \left|\frac{x-a}{x+a}\right|+C\right] \\ \\ &=\log \mid \frac{t+\frac{5}{2}-\frac{1}{2}}{t+\frac{5}{2}-\frac{1}{2}} \mid+C \end{aligned}

$=\log \left|\frac{\frac{2 t+5-1}{2}}{\frac{2 t+5+1}{2}}\right|+C \quad\left[\because u=t+\frac{5}{2}\right]$

.                \begin{aligned} &I=\log \left|\frac{2 t+4}{2 t+6}\right|+C=\log \left|\frac{t+2}{t+3}\right|+C\\ &I=\log \left|\frac{e^{x}+2}{e^{x}+3}\right|+C \end{aligned}$[\therefore t=e^{x}]$