$\int {\mathrm{x}}^2\sqrt{{\mathrm{x}}^6-1}\mathrm{dx}=$

Apply U-substitution : $\mathrm{U}={\mathrm{x}}^6$

$\Rightarrow \int \frac{\sqrt{\mathrm{U}-1}}{6\sqrt{\mathrm{U}}}\mathrm{du}$

$=\frac{1}{6}\cdot \int \frac{\sqrt{\mathrm{u}-1}}{\sqrt{\mathrm{u}}}\mathrm{dv}$

Apply integration by parts: $\mathrm{U}=\sqrt{\mathrm{U}-1}, \mathrm{V}=\frac{1}{\sqrt{\mathrm{U}}}$

$=\frac{1}{6}\left(2\mathrm{U}\sqrt[\frac{1}{2}]{\mathrm{U}-1}-\int \sqrt{\frac{\mathrm{U}}{\mathrm{U}-1}}\mathrm{du}\right)$

$=\frac{1}{6}\left(2{\mathrm{U}}^{\frac{1}{2}}\sqrt{\mathrm{U}-1}-\left(\sqrt{\frac{\mathrm{U}}{\mathrm{U}-1}}(\mathrm{U}-1)+\log |\sqrt{\mathrm{U}}+\sqrt{\mathrm{U}-1}|\right)\right.$

$=\frac{1}{6}\left(2\sqrt{{\mathrm{x}}^6}\sqrt{{\mathrm{x}}^6-1}-\left(\sqrt{\frac{{\mathrm{x}}^6}{{\mathrm{x}}^6-1}}\left({\mathrm{x}}^6-1\right)+\log \sqrt{{\mathrm{x}}^6+\sqrt{{\mathrm{x}}^6}-1}\right)\right)$

$=\frac{1}{6}\left({\mathrm{x}}^3\sqrt{{\mathrm{x}}^6-1}-\log \left({\mathrm{x}}^3+\sqrt{{\mathrm{x}}^6-1}\right)\right)+\mathrm{C}$