Answer
\[\frac{x^2+6}{(x-2)(x^2+16)}=\frac{1}{2(x-2)}+\frac{x+2}{2(x^2+16)}\]
Work Step by Step
In this case the general form of the partial fraction decomposition is
\[\frac{x^2+6}{(x-2)(x^2+16)}=\frac{A}{x-2}+\frac{Bx+C}{x^2+16}\;\;\;...(*)\]
\[x^2+6=A(x^2+16)+(Bx+C)(x-2)\]
\[x^2+6=A(x^2+16)+(Bx^2-2Bx+Cx-2C)\]
\[x^2+6=(A+B)x^2+(C-2B)x+(16A-2C)\]
Comparing like coefficients
\[A+B=1\;\;\;...(1)\]
\[C-2B=0\Rightarrow C=2B\;\;\;...(2)\]
\[16A-2C=6\:\;\;...(3)\]
Using (2) in (3) then (3) becomes
\[16A-4B=6\;\:\;...(4)\]
Multiply (1) by 4 then add to (4)
\[20A=10\Rightarrow A=\frac{1}{2}\]
From (1)
\[B=\frac{1}{2}\]
From (2)
\[C=1\]
From (*)
\[\frac{x^2+6}{(x-2)(x^2+16)}=\frac{1}{2(x-2)}+\frac{\frac{1}{2}x+1}{x^2+16}\]
\[\frac{x^2+6}{(x-2)(x^2+16)}=\frac{1}{2(x-2)}+\frac{x+2}{2(x^2+16)}\]
Hence ,
\[\frac{x^2+6}{(x-2)(x^2+16)}=\frac{1}{2(x-2)}+\frac{x+2}{2(x^2+16)}.\]
