Answer
$ 3\displaystyle \ln|x|+\frac{1}{x}-2\ln|x+3|+C$
Work Step by Step
$x^{3}+3x^{2}=x^{2}(x+3)$
$\displaystyle \frac{x^{2}+8x-3}{x^{3}+3x^{2}}=\frac{A}{x}+\frac{B}{x^{2}}+\frac{C}{x+3}$
$\displaystyle \frac{x^{2}+8x-3}{x^{3}+3x^{2}}=\frac{Ax(x+3)+B(x+3)+Cx^{2}}{x^{2}(x+3)}$
$x^{2}+8x-3=(A+C)x^{2}+(3A+B)x+3B\Rightarrow\left\{\begin{array}{ll}
3B=-3 & \Rightarrow B=-1\\
3A+B=8 & \Rightarrow A=3\\
A+C=1 & \Rightarrow C=-2
\end{array}\right.$
$\displaystyle \int\frac{x^{2}+8x-3}{x^{3}+3x^{2}}dx=\int\left(\frac{3}{x}+\frac{-1}{x^{2}}+\frac{-2}{x+3}\right)dx$
$ = 3\displaystyle \ln|x|-\frac{x^{-1}}{-1}-2\ln|x+3|+C$
$ = 3\displaystyle \ln|x|+\frac{1}{x}-2\ln|x+3|+C$
