Q.$\dfrac{1}{(x-1)(x+2)^{2}}$

Step 1. Set up partial fractions. $\dfrac1{(x-1)(x+2)^2}=\dfrac{A}{x-1}+\dfrac{B}{x+2}+\dfrac{C}{(x+2)^2}$, so $1=A(x+2)^2+B(x-1)(x+2)+C(x-1)$.

Step 2. Solve for $A$. At $x=1$: $1=A(9)\Rightarrow A=\dfrac19$.

Step 3. Solve for $C$. At $x=-2$: $1=C(-3)\Rightarrow C=-\dfrac13$.

Step 4. Solve for $B$. Comparing the coefficient of $x^2$ on both sides (which is $0$ on the left): $0=A+B\Rightarrow B=-\dfrac19$. (Check with the constant term: $A(4)+B(-2)+C(-1)=\tfrac49+\tfrac29+\tfrac13=1$ ✓.) …

• Forgetting the extra $C/(x+2)^2$ term for the repeated factor …