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

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

We have,

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

$=\int \frac{2 e^{2 x}}{\left(e^{2 x}+1\right)^2} d x=\int \frac{1}{\left(e^{2 x}+1\right)^2} d\left(e^{2 x}+1\right)$

$\Rightarrow I =-\frac{1}{e^{2 x}+1}+C=\frac{-e^{-x}}{e^x+e^{-x}}+C$