Question

$\int \frac{2x-1}{(x-1)(x+2)(x-3)}dx$

Answer 1

Let $I=\int \frac{2x-1}{(x-1)(x+2)(x-3)}dxLet\frac{2x-1}{(x-1)(x+2)(x-3)}=\frac{A}{(x-1)}+\frac{B}{(x+2)}+\frac{C}{(x-3)}\Rightarrow 2x-1=A(x+2)(x-3)+B(x-1)(x-3)+C(x-1)(x+2)Putx=3,then6-1=C(3-1)(3+2)\Rightarrow 5=10C\Rightarrow C=\frac{1}{2}Again,putx=1,then2-1=A(1+2)(1-3)\Rightarrow 1=-6A\Rightarrow A=-\frac{1}{6}Now,putx=-2,then-4-1=B(-2-1)(-2-3)\Rightarrow -5=15B\Rightarrow B=-\frac{1}{3}\therefore I=-\frac{1}{6}\int \frac{1}{x-1}dx-\frac{1}{3}\int \frac{1}{x+2}dx+\frac{1}{2}\int \frac{1}{x-3}dx=-\frac{1}{6}log|(x-1)|-\frac{1}{3}|(x+2)|+\frac{1}{2}log|(x-3)|+C$