Answer
\[\frac{3x-2}{(x-5)(x^2+1)}=\frac{1}{2(x-5)}-\frac{x-1}{2(x^2+1)}\]
Work Step by Step
In this case the general form of the partial fraction decomposition is
\[\frac{3x-2}{(x-5)(x^2+1)}=\frac{A}{x-5}+\frac{Bx+C}{x^2+1}\;\;\;...(*)\]
\[3x-2=A(x^2+1)+(Bx+C)(x-5)\]
\[3x-2=A(x^2+1)+(Bx^2-5Bx+Cx-5C)\]
\[3x-2=(A+B)x^2+(C-5B)x+(A-5C)\]
Comparing like coefficients
\[C-5B=3\;\;\;...(1)\]
\[A-5C=-2\;\;\; ...(2)\]
\[A+B=0\Rightarrow B=-A\;\;\;...(3)\]
Using (3) in (1) then (1) becomes
\[C+5A=3\;\;\;...(4)\]
Multiply (4) by 5 then add to (1)
\[26A=13\Rightarrow A=\frac{1}{2}\]
From (3)
\[B=\frac{-1}{2}\]
From (1)
\[C=3+5B\Rightarrow C=3-\frac{5}{2}\]
\[\Rightarrow C=\frac{1}{2}\]
From (*)
\[\frac{3x-2}{(x-5)(x^2+1)}=\frac{1}{2(x-5)}+\frac{\frac{-1}{2}x+\frac{1}{2}}{x^2+1}\]
\[\frac{3x-2}{(x-5)(x^2+1)}=\frac{1}{2(x-5)}-\frac{x-1}{2(x^2+1)}\]
Hence,\[\frac{3x-2}{(x-5)(x^2+1)}=\frac{1}{2(x-5)}-\frac{x-1}{2(x^2+1)}.\]
