Answer
\[\frac{7x+2}{(x-2)(x+2)^2}=\frac{1}{(x-2)}-\frac{1}{x+2}+\frac{3}{(x+2)^2}\]
Work Step by Step
In this case the general form of the partial fraction decomposition is
\[\frac{7x+2}{(x-2)(x+2)^2}=\frac{A}{(x-2)}+\frac{B}{x+2}+\frac{C}{(x+2)^2}\;\;\;...(*)\]
\[7x+2=A(x+2)^2+B(x-2)(x+2)+C(x-2)\]
\[7x+2=A(x^2+4+4x)+B(x^2-4)+C(x-2)\]
\[7x+2=(A+B)x^2+(4A+C)x+(4A-4B-2C)\]
Comparing like coefficients
\[A+B=0\Rightarrow B=-A\;\;\;...(1)\]
\[4A+C=7\;\;\;...(2)\]
\[4A-4B-2C=2\;\;\;...(3)\]
Using (1) in (3) then (3) becomes
\[8A-2C=2\;\;\;...(4)\]
Multiply (2) by 2 then add to (4)
\[16A=16\Rightarrow A=1\]
From (1)
\[B=-1\]
From (2)
\[C=3\]
From (*)
\[\frac{7x+2}{(x-2)(x+2)^2}=\frac{1}{(x-2)}+\frac{-1}{x+2}+\frac{3}{(x+2)^2}\]
Hence,
\[\frac{7x+2}{(x-2)(x+2)^2}=\frac{1}{(x-2)}-\frac{1}{x+2}+\frac{3}{(x+2)^2}.\]
