Answer
$\dfrac{x^3+2x^2-3}{(x^2-2)^2}=\dfrac{x+2}{x^2-2}+\dfrac{2x+1}{(x^2-2)^2}$
Work Step by Step
We will write the partial decomposition as:
$\dfrac{x^3+2x^2-3}{(x^2-2)^2}=\dfrac{Ax+B}{x^2-2}+\dfrac{Cx+D}{(x^2-2)^2} ~~~~(a)$
This implies that $x^3+2x^2-3=(Ax+B)(x^2-2)+(Cx+D) ~~~(b)$
On comparing coefficients, we get $A=1$ and $B=2$
Now, plug $A=1$ and $B=2$ in equation (b) to obtain: $C=2$
and $D=1$
Back substitution $A=1; C=-3$ and $x=-1$ in equation (b).
So, the equation (a) becomes:
$\dfrac{x^3+2x^2-3}{(x^2-2)^2}=\dfrac{x+2}{x^2-2}+\dfrac{2x+1}{(x^2-2)^2}$
