Answer
\[\frac{10}{(x-1)(x^2+9)}=\frac{1}{(x-1)}-\frac{x+1}{x^2+9}\]
Work Step by Step
In this case the general form of the partial fraction decomposition is
\[\frac{10}{(x-1)(x^2+9)}=\frac{A}{(x-1)}+\frac{Bx+C}{x^2+9}\;\;\;...(*)\]
\[10=A(x^2+9)+(Bx+C)(x-1)\]
\[10=A(x^2+9)+(Bx^2-Bx+Cx-C)\]
Comparing like coefficients
\[A+B=0\;\;\;...(1)\]
\[C-B=0\Rightarrow C=B\;\;\;...(2)\]
\[9A-C=10\;\;\;...(3)\]
Using (2) in (3)
\[9A-B=10\;\;\;...(4)\]
Add (1) and (4)
\[10A=10\Rightarrow A=1\]
From (1)
\[B=-1\]
From (2)
\[C=-1\]
From (*)
\[\frac{10}{(x-1)(x^2+9)}=\frac{1}{(x-1)}+\frac{-1x-1}{x^2+9}\]
\[\frac{10}{(x-1)(x^2+9)}=\frac{1}{(x-1)}-\frac{x+1}{x^2+9}\]
Hence
\[\frac{10}{(x-1)(x^2+9)}=\frac{1}{(x-1)}-\frac{x+1}{x^2+9}.\]
