Integral $\int \operatorname{sech}^4 x \, dx$ How we can solve this?$\newcommand{\sech}{\operatorname{sech}}$
$$
\int \sech^4 x \, dx.
$$
I know we can solve the simple case 
$$
\int \sech \, dx=\int\frac{dx}{\cosh x}=\int\frac{dx\cosh x}{\cosh ^2x}=\int\frac{d(\sinh x)}{1+\sinh^2 x}=\int \frac{du}{1+u^2}=\tan^{-1}\sinh x+C.
$$
I am stuck with the $\sech^4$ though. Thank you

Note that
$$
\int \DeclareMathOperator{sech}{sech}{\sech}^4x\,dx=\int{\sech}^2{x}\cdot(1-\tanh^2x)\,dx
$$
Letting $u=\tanh x$ gives $du={\sech}^2x$ so
$$
\int{\sech}^4x\,dx=\int(1-u^2)\,du=u-\frac{u^3}{3}+C=\tanh x-\frac{1}{3}\tanh^3x+C
$$
