815. Calculate (1) $\int \frac{d x}{\sqrt{1+e^{x}}}$
(2) $\int \frac{x^{2}+1}{x \sqrt{x^{4}+1}} d x$.
$$
\begin{array}{l}
\text { (1) } \int \frac{d x}{\sqrt{1+e^{x}}}=\int \frac{d x}{e^{\frac{x}{3}} \sqrt{e^{-x}+1}} \\
=-2 \int \frac{e^{\cdots x}}{\left(e^{-\frac{x}{2}}\right)^{2}+1} \\
=-2 \operatorname{sh}^{-1}\left(e^{-\frac{x}{2}}\right)+C \\
=x-2 \ln \left(1+\sqrt{1+e^{x}}\right)+C . \\
\end{array}
$$
$$
\begin{array}{l}
\text { (2) } \int \frac{x^{2}+1}{x \sqrt{x^{4}+1}} d x=\int\left(\frac{x}{\sqrt{x^{4}+1}}\right. \\
\left.+\frac{1}{x^{3} \sqrt{1+\frac{1}{x^{4}}}}\right) d x \\
\end{array}
$$
$$
\begin{array}{l}
=\frac{1}{2} \int \frac{d\left(x^{2}\right)}{\sqrt{\left(x^{2}\right)^{2}+1}}-\frac{1}{2} \int \frac{d\left(\frac{1}{x^{2}}\right)}{\sqrt{1+\left(\frac{1}{x^{2}}\right)^{2}}} \\
=\frac{1}{2} \operatorname{sh}^{-1}\left(x^{2}\right)-\frac{1}{2} \operatorname{sh}^{-1}\left(\frac{1}{x^{2}}\right)+C \\
=\ln \frac{x^{2}+1+\sqrt{1+x^{4}}}{|x|}+C_{1} .
\end{array}
$$

The simplification of the inverse hyperbolic functions can be easily obtained from the definition through simple transformations.
