Question

Tìm nguyên hàm sau:

6) \(\int \frac{x^2-4x+2}{x^2+2x-3}dx\)

Answer 1

\(\dfrac{x^2-4x+2}{x^2+2x-3}\)

\(=\dfrac{x^2+2x-3-6x-5}{x^2+2x-3}\)

\(=1-\dfrac{6x+5}{\left(x+3\right)\left(x-1\right)}\)

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: \(\dfrac{x^2-4x+2}{x^2+2x-3}=1-\dfrac{13}{4x+12}-\dfrac{11}{4x-4}\)

\(\int\dfrac{x^2-4x+2}{x^2+2x-3}dx=\int1-\dfrac{13}{4x+12}-\dfrac{11}{4x-4}dx\)

\(=x-\dfrac{13}{4}\cdot ln\left|x+3\right|-\dfrac{11}{4}\cdot ln\left|x-1\right|\)