Question

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

Answer 1

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

$=\frac{A}{x+2}+\frac{Bx+C}{x^2+1}$

$\therefore x^2-3=A(x^2+1)+(Bx+C)(x+2)$

$\Rightarrow x^2-3=A{x}^2+A+B{x}^2+2Bx+Cx+2C$

Comparing the coefficients of $x^2,x$ and constant terms, we have

$1=A+B..\left(1\right)$

$0=2B+C...\left(2\right)$

$-3=A+2C...\left(3\right)$

Subtracting equation $\left(1\right)$ from $\left(3\right)$, we have

$2C-B=-3-1=-4...\left(4\right)$

Multiplying equation $\left(4\right)$ by $2$ and adding that to equation $\left(2\right)$, we have

$4C-2B+2B+C=-8$

$\therefore C=-\frac{8}{5}$

$\therefore A=-3-2C=\frac{1}{5},B=1-A=\frac{4}{5}$

$\therefore \frac{x^2-3}{(x+2)(x^2+1)}=\frac{1}{5(x+2)}+\frac{4x-8}{5(x^2+1)}$