Answer
$\tan^{-1}(\tanh t)+c $
Work Step by Step
Since $ u=\tanh t $, then $ du=\operatorname{sech}^2 tdt $.
Now, we have
$$\int \frac{dt}{\cosh^2 +\sinh^2t }=\int \frac{\operatorname{sech}^2tdt}{1 +\tanh^2t }=\int \frac{1}{1+u^2}du\\
=\tan^{-1}u+c=\tan^{-1}(\tanh t)+c.$$
