Question

Resolve $\frac{23x-11x^2}{(2x-1)(9-x^2)}$ into partial fractions.

Answer 1

Assume $\frac{23x-11x^2}{(2x-1)(9-x^2)}=\frac{A}{2x-1}+\frac{B}{3+x}+\frac{C}{3-x}....\left(1\right)$
$\therefore$ $23x-11x^2=A(3+x)(3-x)+B(2x-1)(3-x)+C(2x-1)(3+x)$

$\therefore$ $23x-11x^2=A(9-x^2)+B(7x-2x^2-3)+C(5x+2x^2-3)$

Equating the coefficients of like terms, we get

$2C-2B-A=-11$ ......... $\left(2\right)$

$7B+5C=23$ ......... $\left(3\right)$ and

$9A-3B-3C=0$ ......... $\left(4\right)$

Solving above equations, we get
$A=1,B=4,C=-1$
$\therefore \frac{23x-11x^2}{(2x-1)(9-x^2)}=\frac{1}{2x-1}+\frac{4}{3+x}+\frac{-1}{3-x}$