Answer
\[\frac{4}{x+3}+\frac{1}{3(x-1)}-\frac{10}{3(x+2)}\]
Work Step by Step
In this case the general form of the partial fraction decomposition is
\[\frac{x^2-x+4}{(x + 3)(x - 1)(x+2)}=\frac{A}{x+3}+\frac{B}{x-1}+\frac{C}{x+2}\;\;\;...(*)\]
\[\Rightarrow x^2-x+4=A(x-1)(x+2)+B(x+3)(x+2)+C(x+3)(x-1) \;\;\;...(1)\]
Put $x=-3$ in (1)
\[9+3+4=A(-4)(-1)\]
\[\Rightarrow 16=4A\Rightarrow A=4\]
Put $x=1$ in (1)
\[1-1+4=B(4)(3)\]
\[\Rightarrow B=\frac{4}{12}=\frac{1}{3}\]
Put $x=-2 $ in (1)
\[4+2+4=C(1)(-3)\]
\[\Rightarrow C=\frac{-10}{3}\]
From (*)
\[\frac{x^2-x+4}{(x + 3)(x - 1)(x+2)}=\frac{4}{x+3}+\frac{1}{3(x-1)}+\frac{-10}{3(x+2)}\]
Hence ,\[\frac{x^2-x+4}{(x + 3)(x - 1)(x+2)}=\frac{4}{x+3}+\frac{1}{3(x-1)}+\frac{-10}{3(x+2)}.\]
