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