Question

$\int \frac{6x^2-17x-5}{(x-3)(x-2{)}^2}dx=$

Answer 1

$\int \frac{{6x}^2-17x-5}{(x-3){(x-2)}^2}dx$

Let $\frac{{6x}^2-17x-5}{(x-3){(x-2)}^2}=\frac{A}{x-3}+\frac{B}{(x-2)}+\frac{C}{{(x-2)}^2}$

$\Rightarrow {6x}^2-17x-5=A{(x-2)}^2+B(x-3)(x-2)+C(x-3)$

$=A(x^2-4x+4)+B(x^2-5x+6)+C(x-3)$

Comparing Coefficients,

$A+B=6$

$-4A-5B+C=-17$

$4A+6B-3C=-5$

$\Rightarrow A=-2,B=8,C=15$

$\Rightarrow \int \frac{{6x}^2-17x-5}{(x-3){(x-2)}^2}dx=\int (\frac{-2}{x-3}+\frac{8}{x-2}+\frac{15}{{(x-2)}^2})dx$

$=-2log|x-3|+8log|x-2|-\frac{15}{(x-2)}+C$