Answer
$$\tan e^x-e^x +c$$
Work Step by Step
Given
$$\int e^x\tan^2(e^x)dx $$
Let $$ u =e^x\ \ \ \to \ \ \ du=e^xdx $$
Then
\begin{align*}
\int e^x\tan^2(e^x)dx &= \int \tan^2udu\\
&= \int (\sec^2u-1)du\\
&=\tan u-u +c\\
&= \tan e^x-e^x +c
\end{align*}