#### Provide Solution For  R.D.Sharma Maths Class 12 Chapter 18  Indefinite Integrals Exercise Multiple Choice Questions Question 21 Maths Textbook Solution.

$\frac{-e^{-x}}{e^{x}+e^{-x}}+C$

Given:

$\int \frac{2}{\left(e^{x}+e^{-x}\right)^{2}} d x$

Hint:

You must know about the $\int x^{n} d x$

Explanation:

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

\begin{aligned} &I=\int \frac{d z}{z^{2}} \\ &=\int z^{-2} d z \\ &=\frac{z^{-2+1}}{-2+1}+C \quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+C\right] \\ &=-\frac{1}{z}+C \\ &=\frac{-1}{\left(t^{2}+1\right)}+C \quad\left[\because z=t^{2}+1\right] \\ &=\frac{-1}{e^{2 x}+1}+C \quad\left[\because t=e^{x}\right] \\ &=\frac{-1}{e^{x}\left(e^{x}+\frac{1}{e^{x}}\right)}+C \\ &=\frac{-e^{-x}}{\left(e^{x}+e^{-x}\right)}+C \end{aligned}

Hence $I=\frac{-e^{-x}}{e^{x}+e^{-x}}+C$

Option (a) is correct