Answer
$$\ln \left|\sec e^{t}+\tan e^{t}\right|+C $$
Work Step by Step
We integrate as follows:
Let $ u=e^{t}\ \ \Rightarrow \ \ du=e^{t}dt $, then
\begin{align*}
\int e^{t} \sqrt{\tan ^{2} e^{t}+1} d t&=\int \sqrt{\tan ^{2}u+1} d t\\
&=\int \sec u d t\\
&=\ln \left|\sec u+\tan u\right|+C\\
&=\ln \left|\sec e^{t}+\tan e^{t}\right|+C
\end{align*}
