Answer
$-\dfrac{2}{x-4}+\dfrac{3}{(x-4)^2}$
Work Step by Step
We will write the partial decomposition as:
$\dfrac{11-2x}{(x-4)^2}=\dfrac{A}{x-4}+\dfrac{B}{(x-4)^2} ~~~~(a)$
This implies that $11-2x=A(x-4)+B$
Plug $x=1$ to obtain: $A=-2$
Now, plug $x=4$ to obtain: $B=3$
So, the equation (a) becomes:
$\dfrac{11-2x}{(x-4)^2}=-\dfrac{2}{x-4}+\dfrac{3}{(x-4)^2} $
