Answer
\[\frac{-1}{2(x+4)}+\frac{1}{6(x-2)}+\frac{1}{3(x+1)}\]
Work Step by Step
In this case the general form of the partial fraction decomposition is
\[\frac{2x-1}{(x + 4)(x - 2)(x+1)}=\frac{A}{x+4}+\frac{B}{x-2}+\frac{C}{x+1}\;\;\;...(*)\]
\[\Rightarrow 2x-1=A(x-2)(x+1)+B(x+4)(x+1)+C(x+4)(x-2) \;\;\;...(1)\]
Put $x=-4$ in (1)
\[-9=A(-6)(-3)\Rightarrow A=\frac{-9}{18}\]
\[\Rightarrow A=\frac{-1}{2}\]
Put $x=2$ in (1)
\[4-1=3=B(6)(3)\Rightarrow B=\frac{1}{6}\]
Put $x=-1$ in (1)
\[-3=C(3)(-3)\Rightarrow C=\frac{1}{3}\]
From (*)
\[\frac{2x-1}{(x + 4)(x - 2)(x+1)}=\frac{-1}{2(x+4)}+\frac{1}{6(x-2)}+\frac{1}{3(x+1)}\]
Hence ,
\[\frac{2x-1}{(x + 4)(x - 2)(x+1)}=\frac{-1}{2(x+4)}+\frac{1}{6(x-2)}+\frac{1}{3(x+1)}.\]
