Answer
$$\tan^{-1}\left(\tanh ^{-1} t\right)+C$$
Work Step by Step
Given $$\int \frac{d t}{\cosh ^{2} t+\sinh ^{2} t}$$
Let
$$ u=\tanh t\ \ \ \ \ \ du=\operatorname{sech}^{2} t d t $$
Then
\begin{align*}
\int \frac{d t}{\cosh ^{2} t+\sinh ^{2} t}&=\int \frac{d t}{\cosh ^{2} t(1+\tanh ^{2}t) }\\
&=\int \frac{\operatorname{sech}^{2} t d t}{1+\tanh ^{2} t}\\
&=\int \frac{d u}{1+u^{2}}\\
&=\tan^{-1} u+C\\
&=\tan^{-1}\left(\tanh ^{-1} t\right)+C
\end{align*}
