삼각치환

The integrand has a quadratic polynomial under a square root, which makes us think of trig substitutions. However, it's good practice to look for simpler methods before we jump into more complicated ones, and in this case we find something nicer than a trig substitution: the substitution $u=x^2-1$, $\mathrm{d}u=2x\,\mathrm{d}x$. Then $x\mathrm{d}x = \frac{1}{2}\mathrm{d}u$, and $x^2 ={u+1}$. When $x=\sqrt{10}$, $u=9$, and when $x=\sqrt{17}$, $u=16$.

$$\begin{aligned} \int_{\sqrt{10}}^{\sqrt{17}} \frac{x^3}{\sqrt{x^2-1}}\, \mathrm{d}x&= \int_{\sqrt{10}}^{\sqrt{17}} \frac{x^2}{\sqrt{x^2-1}}\, \cdot x\mathrm{d}x \\&=\frac{1}{2} \int_{9}^{16} \frac{u+1}{\sqrt{u}}\,\mathrm{d}u\\ &=\frac{1}{2} \int_{9}^{16}\left(u^{1/2}+u^{-1/2}\right)\,\mathrm{d}u \\&=\frac{1}{2}\left[\frac{2}{3}u^{3/2} + 2u^{1/2} \right]_{9}^{16} \\&=\frac{1}{2}\left[\frac{2}{3}\cdot 4^3 + 2\cdot 4 -\frac{2}{3}\cdot 3^3 -2\cdot 3 \right]\\ &=\frac{40}{3} \end{aligned}$$