Question

Tính tích phân bất định hàm số hữu tỉ sau :

\(I=\int x^2\left(2-3x^2\right)^8dx\)

Answer 1

Đặt \(t=2-3x^2\)\(\Rightarrow\begin{cases}dt=-6xdx\\x^2=\frac{2-t}{3}\end{cases}\)\(\Leftrightarrow x^2\left(2-3x^2\right)^8=\left(\frac{2-t}{3}\right)t^8=\frac{1}{3}\left(2t^8-t^9\right)\)

Vậy : 

\(I=\int x^2\left(2-3x^2\right)^8dx=\frac{1}{3}\left(2\int t^8dt-\int t^9dt\right)=\frac{2}{27}t^9-\frac{1}{30}t^{10}+C\)

  \(=\frac{2}{27}\left(2-3x^2\right)^9-\frac{1}{30}\left(2-3x^2\right)^{10}+C\)