\(\dfrac{x^2-4x+2}{x^2+2x-3}\)
\(=\dfrac{x^2+2x-3-6x-5}{x^2+2x-3}\)
\(=1-\dfrac{6x+5}{\left(x+3\right)\left(x-1\right)}\)
Đặt \(\dfrac{6x+5}{\left(x+3\right)\left(x-1\right)}=\dfrac{A}{x+3}+\dfrac{B}{x-1}\)
=>\(6x+5=A\left(x-1\right)+B\left(x+3\right)\)
=>\(6x+5=x\left(A+B\right)-A+3B\)
=>\(\left\{{}\begin{matrix}A+B=6\\-A+3B=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}B=\dfrac{11}{4}\\A=6-\dfrac{11}{4}=\dfrac{13}{4}\end{matrix}\right.\)
vậy: \(\dfrac{x^2-4x+2}{x^2+2x-3}=1-\dfrac{13}{4x+12}-\dfrac{11}{4x-4}\)
\(\int\dfrac{x^2-4x+2}{x^2+2x-3}dx=\int1-\dfrac{13}{4x+12}-\dfrac{11}{4x-4}dx\)
\(=x-\dfrac{13}{4}\cdot ln\left|x+3\right|-\dfrac{11}{4}\cdot ln\left|x-1\right|\)
