Question

Integrate the following function defined over
$\frac{x^3-6x^2+10x-2}{x^2-5x+6}$

Answer 1

$\frac{x^3-6x^2+10x-2}{x^2-5x+6}$, on dividing we get

$\frac{x^3-6x^2+10x-2}{x^2-5x+6}=x-1+\frac{-x+4}{x^2-5x+6}$

Now let $\frac{-x+4}{x^2-5x+6}=\frac{-x+4}{(x-2)(x-3)}=\frac{A}{x-2}+\frac{B}{x-3}$

$\Rightarrow \frac{-x+4}{(x-2)(x-3)}=\frac{A(x-3)+B(x-2)}{(x-2)(x-3)}$

$\Rightarrow A(x-3)+B(x-2)=-x+4$

$\Rightarrow A+B=-1$

$-3A-2B=4$

$\Rightarrow 3A+3B=-3$

$-3A-2B=4$

$B=1\Rightarrow A=-2$

$\therefore \frac{-x+4}{x^2-5x+6}=\frac{-2}{x-2}+\frac{1}{x-3}$

$\therefore \int \frac{x^3-6x^2+10x-2}{x^2-5x+6}=\int (x-1+\frac{1}{x-3}-\frac{2}{x-2})dx$

$=\frac{x^2}{2}-x+log(x-3)-2log(x-2)+C$

$=\frac{x^2}{2}-x+log\left(\frac{x-3}{(x-2{)}^2}\right)+C$