Answer
\[\frac{7-2x^2}{(x-1)(x^2+4)}=\frac{1}{(x-1)}-\frac{3(x+1)}{x^2+4}\]
Work Step by Step
In this case the general form of the partial fraction decomposition is
\[\frac{7-2x^2}{(x-1)(x^2+4)}=\frac{A}{(x-1)}+\frac{Bx+C}{x^2+4}\;\;\;...(*)\]
\[7-2x^2=A(x^2+4)+(Bx+C)(x-1)\]
\[7-2x^2=A(x^2+4)+(Bx^2-Bx+Cx-C)\]
\[7-2x^2=(A+B)x^2+(C-B)x+(4A-C)\]
Compare like coefficients both side
\[A+B=-2\;\;\;...(1)\]
\[C-B=0\Rightarrow C=B\;\;\;...(2)\]
\[4A-C=7\;\;\;...(3)\]
Using (2) in (3) then (3) becomes
\[4A-B=7\;\;\;...(4)\]
Add (1) and (4)
\[5A=5\Rightarrow A=1\]
From (1)
\[B=-3\]
From (2)
\[C=-3\]
From (*)
\[\frac{7-2x^2}{(x-1)(x^2+4)}=\frac{1}{(x-1)}+\frac{-3x-3}{x^2+4}\]
\[\frac{7-2x^2}{(x-1)(x^2+4)}=\frac{1}{(x-1)}+\frac{-3(x+1)}{x^2+4}\]
Hence,
\[\frac{7-2x^2}{(x-1)(x^2+4)}=\frac{1}{(x-1)}+\frac{-3(x+1)}{x^2+4}.\]
