Finding Maclaurin series of a natural log function

[tmt1]

I need to find the Maclaurin series of this function:

$$f(x) = ln(1 - x^2)$$

I know that $ln(1 + x)$ equals

$$\sum_{n = 1}^{\infty}\frac{(-1)^{n - 1} x^n}{n}$$

Or, $x - \frac{x^2}{2} + \frac{x^3}{3} ...$

If I swap in $-x^2$ for x, I get:

$$-x^2 + \frac{x^4}{2} - \frac{x^5}{3} + \frac{x^6}{4} ...$$

Is this a valid method?

However, the answer in the text says

$$- (x^2 + \frac{x^4}{2} + \frac{x^6}{3} ...)$$

 

[Prove It]

I need to find the Maclaurin series of this function:

$$f(x) = ln(1 - x^2)$$

I know that $ln(1 + x)$ equals

$$\sum_{n = 1}^{\infty}\frac{(-1)^{n - 1} x^n}{n}$$

Or, $x - \frac{x^2}{2} + \frac{x^3}{3} ...$

If I swap in $-x^2$ for x, I get:

$$-x^2 + \frac{x^4}{2} - \frac{x^5}{3} + \frac{x^6}{4} ...$$

Is this a valid method?

However, the answer in the text says

$$- (x^2 + \frac{x^4}{2} + \frac{x^6}{3} ...)$$

$\displaystyle \begin{align*} \ln{ \left( 1 - x^2 \right) } &= \ln{ \left[ \left( 1 - x \right) \left( 1 + x \right) \right] } \\ &= \ln{ \left( 1 - x \right) } + \ln{ \left( 1 + x \right) } \end{align*}$

so that means

$\displaystyle \begin{align*} \frac{\mathrm{d}}{\mathrm{d}x} \,\left[ \ln{ \left( 1 - x^2 \right) } \right] &= \frac{\mathrm{d}}{\mathrm{d}x}\,\left[ \ln{ \left( 1 - x \right) } + \ln{\left( 1 + x \right) } \right] \\ &= -\frac{1}{1 - x} + \frac{1}{1 + x} \end{align*}$

Can you relate each of these to a geometric series?