Nguyên hàm 6

Tính nguyên hàm $I= \displaystyle\int \dfrac{(x-1)\ln^2 x }{(x\ln x-x+1)^2}dx$

 Để ý $(x\ln x-x+1)'=\ln x$. Dùng phương pháp từng phần ta có
$\begin{aligned}I&=(x-1)\ln x. \left( \dfrac{-1}{x\ln x-x+1}\right)+ \displaystyle\int \dfrac{x\ln x+x-1 }{x(x\ln x-x+1)}dx\\
&=(x-1)\ln x. \left( \dfrac{-1}{x\ln x-x+1}\right)+ \displaystyle\int \dfrac{2x\ln x-(x\ln x-x+1)}{x(x\ln x-x+1)}dx\\
&= (x-1)\ln x. \left( \dfrac{-1}{x\ln x-x+1}\right)+ \displaystyle\int \dfrac{2\ln x}{x\ln x-x+1}dx- \displaystyle\int \dfrac{1}{x}dx\\
&=(x-1)\ln x. \left( \dfrac{-1}{x\ln x-x+1}\right)+ 2\ln |x\ln x-x+1|- \ln |x|+C \end{aligned}$