Đặt \(t=\left(2-3x^{2} \right), \) ta có: \({\rm d}t=-6x.{\rm d}x\Rightarrow x.{\rm d}x=\frac{-1}{6} {\rm d}t\) và \(x^{2} =\frac{2-t}{3} .
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Khi đó ta có:
\(\int x^{3} \left(2-3x^{2} \right)^{8} {\rm d}x =\int x^{2} \left(2-3x^{2} \right)^{8} x{\rm d}x \)
\(=\int \left(\frac{2-t}{3} \right) t^{8} \left(\frac{-1}{6} \right){\rm d}t=\frac{-1}{18} \int \left(2t^{8} -t^{9} \right){\rm d}t \)
\(=\frac{-1}{18} \left(\frac{2t^{9} }{9} -\frac{t^{10} }{10} \right)+C=\frac{-1}{18} \left[\frac{2\left(2-3x^{2} \right)^{9} }{9} -\frac{\left(2-3x^{2} \right)^{10} }{10} \right]+C\)
\(=\frac{-\left(2-3x^{2} \right)^{9} }{81} +\frac{\left(2-3x^{2} \right)^{10} }{180} +C\)