Let $l=\int \log \left(a^2+x^2\right) d x$
By using method of integration by parts, we get
$=\log \left(a^2+x^2\right) \int d x-\int\left\{\frac{d\left(\log \left(a^2+x^2\right)\right.}{d x} \int d x\right\} d x+C$
$\begin{aligned}
& =\log \left(a^2+x^2\right) \int \frac{2 x}{x^2+a^2} \cdot x d x+C \\
& =x \log \left(a^2+x^2\right)-2 \int \frac{x^2}{x^2+a^2} d x+C \\
& =x \log \left(a^2+x^2\right)-2 \int \frac{x^2+a^2}{x^2+a^2} d x+2 \int \frac{a^2}{x^2+a^2} d x \\
& =x \log \left(a^2+x^2\right)-\frac{2 a^2}{a} \tan ^{-1} \frac{x}{a}+C \\
& l=x \log \left(a^2+x^2\right)-2 x+2 a \tan ^{-1} \frac{x}{a}+C
\end{aligned}$
Comparing the above equation with $\int \log \left(x^2+a^2\right) d x=$ $h(x)+\mathrm{C}$, we get
$h(x)=x \log \left(a^2+x^2\right)-2 x+2 a \tan ^{-1}\left(\frac{x}{a}\right)$