Example 4. Calculate $\int \sqrt{x^{2}+a^{2}} d x .(a>0)$
Let $x=a \operatorname{sh} u$, then $\sqrt{x^{2}+a^{2}}=a c h u$, $d x=a c h u d u$,
$$
\begin{array}{l}
=\frac{a^{2}}{2} \ln\left(x+\sqrt{x^{2}+a^{2}}\right)+\frac{x}{2} \sqrt{x^{2}+a^{2}} \\
+C_{1} \text {. } \\
\end{array}
$$

where $c h^{2} u=\frac{1+c h 2 u}{2}$ can be immediately obtained from $c h u$ ～cosu in equation (21).

If this problem is solved using trigonometric substitution, the calculation of the integral is more difficult.
If this problem is solved using integration by parts, it is simpler.
$$
\begin{array}{l} 
\because \quad \int \sqrt{x^{2}+a^{2}} d x=x \sqrt{x^{2}+a^{2}} \\
-\int \frac{x^{2} d x}{\sqrt{x^{2}+a^{2}}} \\
=x \sqrt{x^{2}+a^{2}}-\int \sqrt{x^{2}+a^{2}} d x \\
+a^{2} \int \frac{d x}{\sqrt{x^{2}+a^{2}}}, \\
\int \sqrt{x^{2}+a^{2}} d x=\frac{x}{2} \sqrt{x^{2}+a^{2}} \\
\quad+\frac{a^{2}}{2} \operatorname{sh}^{-1} \frac{x}{a}+C \\
=\frac{x}{2} \sqrt{x^{2}+a^{2}}+\frac{a^{2}}{2} \ln \left(x+\sqrt{x^{2}+a^{2}}\right) \\
+C_{1}
\end{array}
$$