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