How to integrate that function $$\int \frac{1}{\cosh^2 x +a^2}dx$$,
What I did was rewrite $$\cosh^2 x = ({\frac {{e}^x+{e}^{-x}}{2}})^2 $$ then $$\int \frac{1}{({\frac {{e}^x+{e}^{-x}}{2}})^2 +a^2}dx=$$ $$\frac{1}{4}\int \frac{1}{e^{2x}+e^{-2x}+4{a}^2 }dx$$, Multiplu by ${e}^x $ $$\frac{1}{4}\int \frac{e^{x}}{e^{x}(e^{2x}+e^{-2x}+4{a}^2) }dx$$, Let's take ${e}^x=t $, it follows $$\frac{1}{4}\int \frac{dt}{t(t^{2}+t^{-2}+4{a}^2) }dx$$, this integral is, $$\frac{1}{4}\ \frac{t}{t(4a^{2}+t^{2}+{t}^{-2}) }$$ Finally, we only need to rewrite this expression taking care ${e}^x=t $
Expand of the comment from lab. \begin{align} &\int \frac{1}{\cosh^2 x +a^2} dx\\ =&\int \frac{\mathrm{sech} ^2x}{1 +a^2\mathrm{sech} ^2x} dx \\ =&\int \frac{1}{1 +a^2(1-\tanh^2x)} d\tanh x \qquad\text{let $y=\tanh x$}\\ =&\int \frac{d y}{(1+a^2)-a^2y^2}\\ =&\frac{1}{a^2}\int \frac{d y}{-y^2+\frac{1+a^2}{a^2}}\qquad \text{let $y=\sqrt{\frac{1+a^2}{a^2}}\mathrm{tanh}\,z$}\\ =&\frac{1}{a^2}\sqrt{\frac{a^2}{1+a^2}} z\\ =&\frac{1}{a^2}\sqrt{\frac{a^2}{1+a^2}}\mathrm{arctanh}\frac{y}{\frac{\sqrt{1+a^2}}{a}} \\ =&\frac{1}{a\sqrt{1+a^2}}\mathrm{arctanh} \frac{ay}{\sqrt{1+a^2}}\\ =&\frac{1}{a\sqrt{1+a^2}}\mathrm{arctanh}\frac{a \tanh x}{\sqrt{1+a^2}} \end{align}