Question

Resolve into partial fractions:
$\frac{2x^3+x^2-x-3}{x(x-1)(2x+3)}$

Answer 1

Resolve into partial fractions

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

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

Now assume,

$\frac{2x-3}{x(x-1)(2x+3)}=\frac{A}{x}+\frac{B}{x-1}+\frac{C}{2x+3}$

After simplifying this, we get

$\frac{2x-3}{x(x-1)(2x+3)}=\frac{A(x-1)(2x+3)+Bx(2x+3)+Cx(x-1)}{x(x-1)(2x+3)}$

$A(x-1)(2x+3)+Bx(2x+3)+Cx(x-1)=2x-3$

Equating the coefficients, we get

$2A+2B+C=0,A+3B-C=0$ and $-3A=-3$

On Solving for $A,B$ and $C$ , we get

$A=1,B=\frac{-1}{5},C=\frac{-8}{5}$

$\therefore$ $\frac{2x^3+x^2-x-3}{x(x-1)(2x+3)}$

$=1+\frac{1}{x}-\frac{1}{5(x-1)}-\frac{8}{5(2x+3)}$