令 $u=1+x^2$，则 $\mathrm{d}u=2x\mathrm{d}x$，且 $x^2=u-1$。于是

$$\int \frac{x^3}{\sqrt{1+x^2}}\mathrm{d}x=\frac{1}{2}\int (u-1)u^{-\frac{1}{2}}\mathrm{d}u.$$

积分得

$$\frac{1}{3}{u}^{\frac{3}{2}}-u^{\frac{1}{2}}+C=\frac{x^2-2}{3}\sqrt{1+x^2}+C.$$