Question

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

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

Answer 1

Answer: (A)

Let $I =\int \frac{2 d x}{\left(e^{x}+e^{-x}\right)^{2}}=\int \frac{2 d x}{e^{2 x}+e^{-2 x}+2}$
$=\int \frac{2 e^{2 x} d x}{e^{4 x}+2 e^{2 x}+1}$
Put $e^{2 x}=t$
$\Rightarrow 2 e^{2 x} d x=d t$
$\therefore I=\int \frac{d t}{t^{2}+2 t+1}=\int \frac{d t}{(t+1)^{2}}=-\frac{1}{t+1}+c$
$=-\frac{1}{e^{2 x}+1}=-\frac{e^{-x}}{e^{x}+e^{-x}}+c$