Answer
$\displaystyle \ln|x|+\frac{1}{3x}+\frac{1}{3\sqrt{6}}\arctan(\frac{x}{\sqrt{6}})+C$
Work Step by Step
$I=\displaystyle \int\frac{x^{3}+6x-2}{x^{4}+6x^{2}}dx$
$\displaystyle \frac{x^{3}+6x-2}{x^{2}(x^{2}+6)}=\frac{A}{x}+\frac{B}{x^{2}}+\frac{Cx+D}{x^{2}+6}$
$x^{3}+6x-2=Ax(x^{2}+6)+B(x^{2}+6)+x^{2}(Cx+D)$
$x^{3}+6x-2=Ax^{3}+6Ax+Bx^{2}+6B+ Cx^{3}+Dx^{2}$
$x^{3}+6x-2=(A+C)x^{3}+(B+D)x^{2} +6Ax+6B$
$6B=-2\Rightarrow B=-1/3$
$6A=6\Rightarrow A=1$
$B+D=0\Rightarrow D=1/3$
$A+C=1\Rightarrow C=0$
$\displaystyle \frac{x^{3}+6x-2}{x^{2}(x^{2}+6)}=\frac{1}{x}+\frac{-1/3}{x^{2}}+\frac{0+1/3}{x^{2}+6}$
$=\displaystyle \frac{1}{x}-\frac{1}{3}\cdot\frac{1}{x^{2}}+\frac{1}{3}\frac{1}{x^{2}+6}$
$I=\displaystyle \ln|x|+\frac{1}{3x}+\frac{1}{3\sqrt{6}}\arctan(\frac{x}{\sqrt{6}})+C$
