Q.Resolve into partial fractions: $\dfrac{x+4}{(x+1)(x-1)^2}$.

A repeated linear factor $(x-1)^2$ needs two terms in the decomposition — one over $(x-1)$ and one over $(x-1)^2$ — alongside the usual term for the simple factor $(x+1)$.

$\dfrac{x+4}{(x+1)(x-1)^2}=\dfrac{A}{x+1}+\dfrac{B}{x-1}+\dfrac{C}{(x-1)^2}$

$x+4=A(x-1)^2+B(x+1)(x-1)+C(x+1)$

At $x=1$: $5=C(2)\Rightarrow C=\dfrac{5}{2}$

At $x=-1$: $3=A(4)\Rightarrow A=\dfrac{3}{4}$

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