#### explain solution RD Sharma class 12 chapter Indefinite Integrals exercise 18.9 question 60 maths

Answer: $1+e^{x}-\log \left|1+e^{x}\right|+c$

Hint: Use substitution method to solve this integral

Given: $\int \frac{e^{2 x}}{1+e^{x}} d x$

Solution:

$I=\int \frac{e^{2 x}}{1+e^{x}} d x$

\begin{aligned} &\text { Put } 1+e^{x}=t \Rightarrow e^{x} d x=d t \Rightarrow d x=\frac{d t}{e^{x}} \\ &\text { Then, } I=\int \frac{e^{2 x}}{t} \cdot \frac{d t}{e^{x}}=\int \frac{\left(e^{x}\right)^{2}}{t} \cdot \frac{d t}{e^{x}} \end{aligned}

$=\int \frac{e^{x}}{t} d t=\int \frac{t-1}{t} d t \quad\left[\because 1+e^{x} \Rightarrow e^{x}=t-1\right]$

\begin{aligned} &=\int\left(\frac{t}{t}-\frac{1}{t}\right) d t=\int\left(1-\frac{1}{t}\right) d t \\ &=\int 1 . d t-\int \frac{1}{t} d t=\int t^{0} d t-\int \frac{1}{t} d t \end{aligned}

$=\frac{t^{0+1}}{0+1}-\log |t|+c$                        $\left[\begin{array}{l} \because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c \\ \int \frac{1}{x} d x=\log |x|+c \end{array}\right]$

\begin{aligned} &=t-\log |t|+c \\ &=1+e^{x}-\log \left|1+e^{x}\right|+c\; \; \; \; \; \; \; \; \; \left[\because t=1+e^{x}\right] \end{aligned}