$I=\int\limits x^3\sqrt{x^3+1} dx$Đặt $\left\{ \begin{array}{l} x=u\\ x^2\sqrt{x^3+1}dx=dv \end{array} \right.\Rightarrow \left\{ \begin{array}{l} dx=du\\ \frac{2}{9}( x^3+1)\sqrt{x^3+1}=v \end{array} \right.$
$I=\frac{2}{9}x(x^3+1)\sqrt{x^3+1}-\int\limits \frac{2}{9}(x^3+1)\sqrt{x^3+1}dx$
$I=\frac{2}{9}x(x^3+1)\sqrt{x^3+1}-\frac{2}{9}I-\frac{2}{9}\int\limits \sqrt{x^3+1}dx$
Tịt rồi =))