Answer
$\dfrac{5-2x}{(x-1)(x^2+2)}=\dfrac{1}{x-1}-\dfrac{x+3}{x^2+2} $
Work Step by Step
We will write the partial decomposition as:
$\dfrac{5-2x}{(x-1)(x^2+2)}=\dfrac{A}{x-1}+\dfrac{Bx+C}{x^2+2} ~~~~(a)$
This implies that $5-2x=A(x^2+2)+(Bx+c)(x-1) ~~~(b)$
Plug $x=1$ to obtain: $A=1$
Now, plug $A=1$ in equation (b) to obtain: $C=-3$
Back substitution $A=1; C=-3$ and $x=-1$ in equation (b).
So, the equation (a) becomes:
$\dfrac{5-2x}{(x-1)(x^2+2)}=\dfrac{1}{x-1}+\dfrac{-x-3}{x^2+2} \\ \dfrac{5-2x}{(x-1)(x^2+2)}=\dfrac{1}{x-1}-\dfrac{x+3}{x^2+2} $
