Answer
$\frac{3}{x+2}-\frac{1}{(x+2)^2}-\frac{1}{(x+3)^2}$
Work Step by Step
Step 1. As the denominator has already been factorized, assume the end results as: $\frac{A}{x+2}+\frac{B}{(x+2)^2}+\frac{C}{x+3}+\frac{D}{(x+3)^2}$
Step 2. Combine the above functions as:
$\frac{A(x+2)(x+3)^2+B(x+3)^2+C(x+2)^2(x+3)+D(x+2)^2}{(x+2)^2(x+3)^2}=\frac{(A+C)x^3+(8A+B+7C+D)x^2+(21A+6B+16C+4D)x+(18A+9B+12C+4D)}{(x+2)^2(x+3)^2}$
Step 3. Compare the above result with the original expression to set up the following system of equations:
\begin{cases} A+C=3 \\ 8A+B+7C+D=22\\21A+6B+16C+4D=53\\18A+9B+12C+4D=41 \end{cases}
Step 5. Use substitution $C=3-A$ for last three equations to get:
\begin{cases} A+B+D=1\\5A+6B+4D=5\\6A+9B+4D=5 \end{cases}
Step 6. Take the difference between the last two equations to get $A+3B=0$ or $A=-3B$, use this substitution for the first two equations and solve the remaining variables to get $A=3,B=-1,C=0,D=-1$
Step 7. Write the final results as:
$\frac{3}{x+2}-\frac{1}{(x+2)^2}-\frac{1}{(x+3)^2}$
