Answer
$$\frac{1}{2} e^x \sqrt{e^{2x}-1} -\frac{1}{2} \ln |e^x +\sqrt{e^{2x}-1}|+C$$
Work Step by Step
Given $$\int e^{x} \sqrt{e^{2 x}-1} d x$$
Let
$$e^x= \sec u\ \ \ \ \ \ \ \ e^xdx=\sec u\tan udu $$
Then
\begin{align*}
\int e^{x} \sqrt{e^{2 x}-1} d x&=\int \sqrt{\sec^2 u-1} \sec u\tan udu \\
&=\int \tan^2 u \sec udu\\
&=\int (\sec^2 u-1)\sec udu\\
&=\int (\sec^3 u-\sec u)du
\end{align*}
Use
$$\int \sec ^{3} u d u=\frac{1}{2} \sec u \tan u+\frac{1}{2} \ln |\sec u+\tan u|+C $$
Then
\begin{align*}
\int e^{x} \sqrt{e^{2 x}-1} d x &=\int (\sec^3 u-\sec u)du\\
&=\frac{1}{2} \sec u \tan u-\frac{1}{2} \ln |\sec u+\tan u|+C \\
&=\frac{1}{2} e^x \sqrt{e^{2x}-1} -\frac{1}{2} \ln |e^x +\sqrt{e^{2x}-1}|+C
\end{align*}
