$\int \frac{1}{x^{2}-1} d x$
$=\int_{2}^{1} \frac{(x-1)+1}{(x-1)(x+1)} d x-\int_{2}^{1} \frac{d x}{x+1}+\int_{2}^{1}\left(\frac{1}{x-1}-\frac{1}{x+1}\right) \frac{1}{2} d x$
$=|\log | x+1||_{2}^{1}+\frac{1}{2}\left[\left.\log \left|\frac{x-1}{x+1}\right|\right|_{2} ^{1}\right.$
$=(\log 4-\log 3)+\frac{1}{2}\left|\log \left(\frac{2}{4}\right)-\log \left(\frac{1}{3}\right)\right|$
$=\log \left(\frac{4}{3}\right)+\frac{1}{2} \left|\frac{\log \left(\frac{1}{2}\right)}{1}\right|-\log \left(\frac{4}{3}\right)+\log \left|\frac{3}{2}\right|^{2}$
$=\log \left(\frac{4}{3} x \sqrt{\frac{3}{2}}\right)^{1}-\log \left(\frac{8}{3}\right)^{2} \mid-\frac{1}{2} \log \left(\frac{8}{3}\right)$
This problem can also be solved an follows :
$\begin{array}{l}
\int_{0}^{1} \frac{x}{x^{2}-1} d x \\
-\int_{2}^{1} \frac{x}{(x-1)(x+1)} d x-\frac{1}{2} \int_{0}^{1}\left|\frac{1}{x-1}+\frac{1}{x+1}\right| d x
\end{array}$
$=\frac{1}{2}\left\{[\log (x-1)]_{2}^{3}+[\log (x+1)]_{2}^{3}\right\}$
$=\frac{1}{2}\left\{[\log [(x-1)(x+1)]]_{2}^{3}\right\}=\frac{1}{2}\left[\log \left(x^{2}-1\right)\right]_{2}^{3}$
$=\frac{1}{2}[\log (9-1)-\log (4-1)]=\frac{1}{2} \log \left(\frac{8}{3}\right)$