Partial fraction of $\frac{2x^2-9x-9}{x^3-9x}$ I'm doing some questions from Anton, 8th edition, page 543, question 13. I've found a answer but it does not match with the answer given at the last pages.
Questions asks to solve $\int{\frac{2x^2-9x-9}{x^3-9x}}$
So:
$$\frac{2x^2-9x-9}{x(x^2-9)}$$
then:
$$\frac{2x^2-9x-9}{x(x-3)(x+3)}$$
(using  $a(x-x')(x-x'')$)
Continuing:
$$2x^2-9x-9 = \frac{A}{x} + \frac{B}{(x-3)} + \frac{C}{(x+3)}$$
Solving the product:
$$A(x^2-9)+Bx(x+3)+Cx(x-3)$$
Ended up with a system, because: $(A+B+C)x^2+(3B-3C)x+(-9A)$
\begin{cases} A+B+C =2 \\ 3B-3C = -9 \\ -9A = -9\end{cases}
then:
$A = 1$
$B = -1$
$C = 2$
The answer I found:
$$ ln |x| - ln|x-3| + 2 ln|x+3| + C $$
answer given by author:
$$ ln |\frac{x(x+3)^2}{x-3}| + C$$
What am I missing?

You're answer is fine and is equivalent to the author's.
To see why, recall:
$$\ln a - \ln b = \ln\left(\frac ab\right)$$
$$\ln a + \ln b = \ln(ab)$$
$$2 \ln a = \ln(a^2)$$
