Ta có
$I = \int x^3\sqrt{x^2 + 1}dx$
$= \dfrac{1}{2} \int x^2 \sqrt{x^2 + 1} d(x^2)$
Đặt $t = x^2$. Khi đó
$I = \dfrac{1}{2} \int t \sqrt{t+1} dt$
$= \dfrac{1}{2} \int t . \dfrac{2}{3} d(\sqrt{(t+1)^3})$
$= \dfrac{1}{3} \int t d(\sqrt{(t+1)^3})$
$= \dfrac{1}{3} \left( t.\sqrt{(t+1)^3} - \int \sqrt{(t+1)^3} dt \right)$
$= \dfrac{1}{3} \left( t\sqrt{(t+1)^3} - \dfrac{2}{5} \sqrt{(t+1)^5}+c_1 \right)$
$= \dfrac{t\sqrt{(t+1)^3}}{3} - \dfrac{2}{15} \sqrt{(t+1)^5} + c$
