Question

The value of the integral $\int \frac{1}{e^{2x}+e^{-2x}}dx$ is equal to

• A. $2{\tan }^{-1}(e^{2x})+C$
• B. ${\tan }^{-1}(e^{2x})+C$
• C. $\frac{1}{2}{\tan }^{-1}(e^{2x})+C$
• D. $\frac{-1}{{(e^{2x}+e^{-2x})}^2}+C$

Answer 1

Answer: (C)

$\int \frac{dx}{e^{2x}+e^{-2x}}\Rightarrow \int \frac{e^{2x}}{e^{4x}+1}dx$
$=\int \frac{e^{2x}}{1+{(e^{2x})}^2}dx,$
put $t=e^{2x}$ $dt=2e^{2x}dx$
$\Rightarrow$ $\int \frac{dt}{2(1+t^2)}=\frac{1}{2}{\tan }^{-1}t+C$
$\Rightarrow$ $=\frac{1}{2}{\tan }^{-1}(e^{2x})+C$