Find : \[\int x^{2} \log \left(x^{2}-1\right) d x \]

Integrating by parts (\(\displaystyle \log(x^2-1)\) as first function): \[\begin{aligned} \int x^2\log(x^2-1)\,dx &= \frac{x^3}{3}\log(x^2-1) - \int \frac{2x}{x^2-1}\cdot\frac{x^3}{3}\,dx \\ &= \frac{x^3}{3}\log(x^2-1) - \frac{2}{3}\int \frac{x^4-1+1}{x^2-1}\,dx \\ &= \frac{x^3}{3}\log(x^2-1) - \frac{2}{3}\left[\int(x^2+1)\,dx + \int\frac{dx}{x^2-1}\right] \\ &= \frac{x^3}{3}\log(x^2-1) - \frac{2}{3}\left[\frac{x^3}{3}+x+\frac{1}{2}\log\left|\frac{x-1}{x+1}\right|\right] + C \end{aligned} \]