Question

The value of the integral $\int \frac{dx}{(e^x+e^{-x}{)}^2}$ is

• A. $\frac{1}{2}(e^{2x}+1)+c$
• B. $\frac{1}{2}(e^{-2x}+1)+c$
• C. $-\frac{1}{2}(e^{2x}+1{)}^{-2}+c$
• D. $\frac{1}{4}(e^{2x}-1)+c$

Answer 1

The correct option is C $-\frac{1}{2}(e^{2x}+1{)}^{-2}+c$

$I=\int \frac{dx}{(e^x+e^{-x{)}^2}}=\int \frac{dx}{e^{-2x}(e^{2x}+1{)}^2}$
$=\int \frac{e^{2x}}{(1+e^{2x{)}^2}}dx$
Put $e^x=t\Rightarrow e^{x}dx=dt$
$I=\int \frac{tdt}{(1+t^2{)}^2}=\frac{1}{2}\int \frac{2tdt}{(1+t^2{)}^2}$
$=\frac{1}{2}\left(\begin{array}{c}\frac{-1}{1+t^2}\end{array}\right)+c$
$=\frac{1}{2}\frac{(-1)}{(1+e^{2x}{)}^2}+c$