Answer
\[\frac{7x+4}{(x+1)^3(x-2)}=\frac{-2}{3(x+1)}+\frac{-2}{(x+1)^2}+\frac{1}{(x+1)^3}+\frac{2}{3(x-2)}\]
Work Step by Step
In this case the general form of the partial fraction decomposition is
\[\frac{7x+4}{(x+1)^3(x-2)}=\frac{A}{(x+1)}+\frac{B}{(x+1)^2}+\frac{C}{(x+1)^3}+\frac{D}{x-2}\;\;\;...(*)\]
\[7x+4=A(x+1)^2(x-2)+B(x+1)(x-2)+C(x-2)+D(x+1)^3\]
\[7x+4=A(x^2+2x+1)(x-2)+B(x^2-x-2)+C(x-2)+D(x^3+1+3x^2+3x)\]
\[7x+4=A(x^3-3x-2)+B(x^2-x-2)+C(x-2)+D(x^3+1+3x^2+3x)\]
\[7x+4=(A+D)x^3+(B+3D)x^2+(-3A-B+C+3D)x+(-2A-2B-2C+D)\]
Comparing both sides
\[A+D=0\Rightarrow A=-D\;\;\;...(1)\]
\[B+3D=0\Rightarrow B=-3D\;\;\;...(2)\]
\[-3A-B+C+3D=7\;\;\;...(3)\]
\[-2A-2B-2C+D=4\;\;\;...(4)\]
Using (1) and (2) in (3) then (3) becomes
\[C+9D=7\;\;\;...(5)\]
Using (1) and (2) in (4) then (4) becomes
\[-2C+9D=4\;\;\;...(6)\]
Subtract (6) from (5)
\[3C=3\Rightarrow C=1\]
From (5)
\[D=\frac{2}{3}\]
From (1)
\[A=\frac{-2}{3}\]
From (2)
\[B=-2\]
From (*)
\[\frac{7x+4}{(x+1)^3(x-2)}=\frac{-2}{3(x+1)}+\frac{-2}{(x+1)^2}+\frac{1}{(x+1)^3}+\frac{2}{3(x-2)}\]
Hence ,
\[\frac{7x+4}{(x+1)^3(x-2)}=\frac{-2}{3(x+1)}+\frac{-2}{(x+1)^2}+\frac{1}{(x+1)^3}+\frac{2}{3(x-2)}.\]
