Sử dụng PP tích phân từng phân
$I=\int\limits_{0}^{1}\ln\left(x+1\right)d\left(x^2-x\right)=\left[ {\left(x^2-x\right)\ln\left(x+1\right) } \right]_{0}^{1}-\int\limits_{0}^{1}\frac{x^2-x}{x+1}dx$
$=\left[ {\left(x^2-x\right)\ln\left(x+1\right) } \right]_{0}^{1}-\int\limits_{0}^{1}\left ( x-2+\frac{2}{x+1} \right )dx $
$=\left[ {\left(x^2-x\right)\ln\left(x+1\right) -\frac{1}{2}x^2+2x-2\ln|x+1|} \right]_{0}^{1}$