Đặt \(t=1-x^{3} ,\) ta có: \({\rm d}t=-3x^{2} {\rm d}x\Rightarrow x^{2} {\rm d}x=\frac{-1}{3} {\rm d}t\) và \(x^{3} =1-t.\)
Khi đó: \(\int x^{5} \left(1-x^{3} \right)^{3} {\rm d}x=\int x^{3} \left(1-x^{3} \right)^{3} x^{2} {\rm d}x\)
\(=\int \left(1-t\right)t^{3} \left(\frac{-1}{3} \right){\rm d}t =\frac{-1}{3} \int \left(t^{3} -t^{4} \right){\rm d} t\)
\(=\frac{-1}{3} \left(\frac{t^{4} }{4} -\frac{t^{5} }{5} \right)+C=\frac{-1}{3} \left[\frac{\left(1-x^{3} \right)^{4} }{4} -\frac{\left(1-x^{3} \right)^{5} }{5} \right]+C\)
\(=\frac{-\left(1-x^{3} \right)^{4} }{12} +\frac{\left(1-x^{3} \right)^{5} }{15} +C\)