Question

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