Answer
$3\ln \left|x\right|+\frac{1}{x}-2\ln \left|x+3\right|+C$
Work Step by Step
Given
$$\int \frac{x^2+8x-3}{x^3+3x^2}\:dx $$
Using partial fractions
\begin{aligned}
\frac{x^2+8x-3}{x^2\left(x+3\right)}&=\frac{a_0}{x}+\frac{a_1}{x^2}+\frac{a_2}{x+3}\\
&= \frac{a_0x\left(x+3\right)+a_1\left(x+3\right)+a_2x^2}{x^2\left(x+3\right)}\\
x^2+8x-3&=a_0x\left(x+3\right)+a_1\left(x+3\right)+a_2x^2
\end{aligned}
at $x= 0\ \ \to \ a_1= -1$, $x= -3 \to a_2 =-2$, then
$$x^2+8x-3=a_0x\left(x+3\right)+\left(-1\right)\left(x+3\right)+\left(-2\right)x^2$$
by comparing
$$1\cdot \:x^2+8x-3=x^2\left(a_0-2\right)+x\left(3a_0-1\right)-3$$
we get $a_0= 3$
\begin{aligned}
\int \frac{x^2+8\:x-3}{x^3+3\:x^2}\:d\:x
&=\int \left(\frac{3}{x}-\frac{1}{x^2}-\frac{2}{x+3}\right)dx\\
&= 3\ln \left|x\right|-\left(-\frac{1}{x}\right)-2\ln \left|x+3\right|+C\\
&=3\ln \left|x\right|+\frac{1}{x}-2\ln \left|x+3\right|+C\end{aligned}
