Answer
\[\frac{3x+4}{x^2(x^2+4)}=\frac{3}{4x}+\frac{1}{x^2}-\frac{3x+4}{4(x^2+4)}\]
Work Step by Step
In this case the general form of the partial fraction decomposition is
\[\frac{3x+4}{x^2(x^2+4)}=\frac{A}{x}+\frac{B}{x^2}+\frac{Cx+D}{x^2+4}\;\;\;...(*)\]
\[3x+4=Ax(x^2+4)+B(x^2+4)+(Cx+D)x^2\]
\[3x+4=A(x^3+4x)+B(x^2+4)+(Cx^3+Dx^2)\]
\[3x+4=(A+C)x^3+(B+D)x^2+(4A)x+4B\]
Comparing like coefficents
\[A+C=0\;\;\;...(1)\]
\[B+D=0\;\;\;...(2)\]
\[4A=3\Rightarrow A=\frac{3}{4}\]
\[4B=4\Rightarrow B=1\]
From (1)
\[C=\frac{-3}{4}\]
From (2)
\[D=-1\]
From (*)
\[\frac{3x+4}{x^2(x^2+4)}=\frac{3}{4x}+\frac{1}{x^2}+\frac{\frac{-3}{4}x-1}{x^2+4}\]
\[\frac{3x+4}{x^2(x^2+4)}=\frac{3}{4x}+\frac{1}{x^2}-\frac{3x+4}{4(x^2+4)}\]
Hence ,
\[\frac{3x+4}{x^2(x^2+4)}=\frac{3}{4x}+\frac{1}{x^2}-\frac{3x+4}{4(x^2+4)}.\]
