Partial fraction of $\frac{2x^2-9x-9}{x^3-9x}$

Question

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?

Answer 1

You're answer is fine and is equivalent to the author's.

To see why, recall:

$$\ln a - \ln b = \ln\left(\frac ab\right)$$

$$\ln a + \ln b = \ln(ab)$$

$$2 \ln a = \ln(a^2)$$

Answer 2

Using properties of logarithm ($\ln a + \ln b = \ln (ab)$) you have $$\ln |x| - \ln |x-3| + 2 \ln |x+3| = \ln \frac{|x|(x+3)^2}{|x-3|}$$