Answer
\[\frac{5x^2+3}{(x+1)(x-1)^2}=\frac{2}{(x+1)}+\frac{3}{(x-1)}+\frac{4}{(x-1)^2}\]
Work Step by Step
In this case the general form of the partial fraction decomposition is
\[\frac{5x^2+3}{(x+1)(x-1)^2}=\frac{A}{x+1}+\frac{B}{x-1}+\frac{C}{(x-1)^2}\;\;\;...(*)\]
\[5x^2+3=A(x-1)^2+B(x+1)(x-1)+C(x+1)\]
\[5x^2+3=A(x^2-2x+1)+B(x^2-1)+C(x+1)\]
\[5x^2+3=(A+B)x^2+(-2A+C)x+(A-B+C)\]
Comparing like coefficients
\[A+B=5\;\;\;...(1)\]
\[-2A+C=0\Rightarrow C=2A\;\;\;...(2)\]
\[A-B+C=3\;\;\;...(3)\]
Using (2) then (3) becomes
\[3A-B=3\;\;\;...(4)\]
Add equation (1) and (4)
\[4A=8\Rightarrow A=2\]
From (2)
\[C=4\]
From (4)
\[B=3\]
Then (*) becomes
\[\frac{5x^2+3}{(x+1)(x-1)^2}=\frac{2}{(x+1)}+\frac{3}{(x-1)}+\frac{4}{(x-1)^2}\]
Hence ,
\[\frac{5x^2+3}{(x+1)(x-1)^2}=\frac{2}{(x+1)}+\frac{3}{(x-1)}+\frac{4}{(x-1)^2}.\]
