Answer
$\ln|x|+\frac{1}{3x}+\frac{1}{3\sqrt 6}\tan^{-1}\left(\frac{x}{\sqrt 6}\right)+C$
Work Step by Step
$\int\frac{x^3+6x-2}{x^4+6x^2}dx$ = $\int\frac{x^3+6x-2}{x^2(x^2+6)}dx$ = $\int\left(\frac{A}{x}+\frac{B}{x^2}+\frac{Cx+D}{x^2+6}\right) dx$
$x^3+6x-2$ = $Ax(x^2+6)+B(x^2+6)+x^2(Cx+D)$
$x^3+6x-2$ = $A(x^3+6x)+B(x^2+6)+(Cx^3+Dx^2)$
$1$ = $A+C$
$0$ = $B+D$
$6$ = $6A$
$-2$ = $6B$
$A$ = $1$, $B$ = $-\frac{1}{3}$, $C$ = $0$, $D$ = $\frac{1}{3}$
So
$\int\frac{x^3+6x-2}{x^4+6x^2}dx$ = $\int\left[\frac{1}{x}-\frac{1}{3x^2}+\frac{1}{3(x^2+6)}\right]dx$ = $\ln|x|+\frac{1}{3x}+\frac{1}{3\sqrt 6}\tan^{-1}\left(\frac{x}{\sqrt 6}\right)+C$
