Answer
$$\sqrt{1+e^{2 a}}+\frac{1}{2} \ln \frac{\sqrt{1+e^{2 a}}-1}{\sqrt{1+e^{2 a}}+1}-\sqrt{2}+\ln (1+\sqrt{2})$$
Work Step by Step
Since
$$ \int_{0}^{a} \sqrt{1+e^{2 x}} d x$$
Let
$$u= \sqrt{1+e^{2 x}} \to du= \frac{e^{2 x}}{ \sqrt{1+e^{2 x}}}dx\to dx=\frac{u^2}{u^2-1}du$$
Then
\begin{aligned}
\int_{0}^{a} \sqrt{1+e^{2 x}} d x &=\int_{x=0}^{x=a} u \cdot \frac{u}{u^{2}-1} d u=\int_{x=0}^{x=a} \frac{u^{2}}{u^{2}-1} d u=\int_{x=0}^{x=a}\left(1+\frac{1}{u^{2}-1}\right) d u \\
&=\int_{x=0}^{x=a}\left(1+\frac{1}{2} \frac{1}{u-1}-\frac{1}{2} \frac{1}{u+1}\right) d u=\left.\left(u+\frac{1}{2} \ln (u-1)-\frac{1}{2} \ln (u+1)\right)\right|_{x=0} ^{x=a} \\
&=\left.\left[\sqrt{1+e^{2 x}}+\frac{1}{2} \ln \left(\frac{\sqrt{1+e^{2 x}}-1}{\sqrt{1+e^{2 x}}+1}\right)\right]\right|_{0} ^{a} \\
&=\sqrt{1+e^{2 a}}+\frac{1}{2} \ln \frac{\sqrt{1+e^{2 a}}-1}{\sqrt{1+e^{2 a}}+1}-\sqrt{2}+\frac{1}{2} \ln \frac{1+\sqrt{2}}{\sqrt{2}-1} \\
&=\sqrt{1+e^{2 a}}+\frac{1}{2} \ln \frac{\sqrt{1+e^{2 a}}-1}{\sqrt{1+e^{2 a}}+1}-\sqrt{2}+\ln (1+\sqrt{2})
\end{aligned}
