Đặt \(\left\{{}\begin{matrix}u=ln\left(x+1\right)\\dv=\left(x^2-1\right)dx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=\frac{dx}{x+1}\\v=\frac{1}{3}x^3-x\end{matrix}\right.\)
\(\Rightarrow I=\left(\frac{1}{3}x^3-x\right)ln\left(x+1\right)-\frac{1}{3}\int\frac{\left(x^3-3x\right)}{x+1}dx\)
Xét \(J=\int\frac{\left(x^3-3x\right)dx}{x+1}=\int\left(x^2-x-2+\frac{2}{x+1}\right)dx\)
\(\Rightarrow J=\frac{1}{3}x^3-\frac{1}{2}x^2-2x+2ln\left(x+1\right)\)
\(\Rightarrow I=\left(\frac{1}{3}x^3-x\right)ln\left(x+1\right)-\frac{1}{3}\left(\frac{1}{3}x^3-\frac{1}{2}x^2-2x+2ln\left(x+1\right)\right)+C\)
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