MacLaurin series of $\ln(1-x^2)$ 
The MacLaurin series for $\ln(1 + x)$ is obtained from the series for $\frac{1}{1 + x}$
  by integration. Use this and appropriate substitutions to obtain the MacLaurin series for $\ln(1-x^2)$. 

I did $$\int\frac{2}{x-\frac{1}{x}}$$ but I'm not sure how I'd go about the rest of the substitution in order to find the series.
 A: I think what the question is asking is to use
$$
\log(1+x)=x-\frac{x^2}2+\frac{x^3}3-\frac{x^4}4+\dots\tag{1}
$$
then substitute $x\mapsto-x$
$$
\log(1-x)=-x-\frac{x^2}2-\frac{x^3}3-\frac{x^4}4-\dots\tag{2}
$$
Then using the identity that $\log(ab)=\log(a)+\log(b)$ to get
$$
\begin{align}
\log(1-x^2)
&=-2\frac{x^2}{2}-2\frac{x^4}4-2\frac{x^6}6-\dots\\
&=-x^2-\frac{x^4}2-\frac{x^6}3-\dots\tag{3}
\end{align}
$$
Note than you could also substitute $x\mapsto-x^2$ into $(1)$ to get $(3)$ directly.
A: Well, presumably you already know the series for $\ln(1+x)$. The hint is just telling you to take that series and replace $x$ by $-x^2$.
A: Two solutions, both based on 
$\ln(1+x) = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n x^n}$
for $-1 < x < 1$:


*

*Put $-x^2$ for $x$.
Then
$\ln(1-x^2) = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n (-x^2)^n}
=-\sum_{n=1}^{\infty} \frac{1}{n x^{2n}}
$.

*Put $-x$ for $x$.
Then
$\ln(1-x) = \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n (-x)^n}
=\sum_{n=1}^{\infty} \frac{-1}{n x^{n}}
$.
Add this to the original series, and
$\begin{align}
\ln(1-x^2) = \ln(1-x)+\ln(1+x)
&=\sum_{n=1}^{\infty} (\frac{(-1)^{n-1}}{n x^n}+\frac{-1}{n x^n})\\
&=\sum_{n=1}^{\infty} \frac{(-1)^{n-1}-1}{n x^n}\\
&=\sum_{n=1, n \text{ even}}^{\infty} \frac{(-1)^{n-1}-1}{n x^n}\\
&=\sum_{n=1}^{\infty} \frac{-2}{2n x^{2n}}\\
&=-\sum_{n=1}^{\infty} \frac{1}{n x^{2n}}\\
\end{align}
$
Fortunately, the answers agree.
A: If you wanted to use the same type of derivation used to come up with the Maclaurin series of $\ln (1-x)$ this is how it would be done
\begin{align}
\frac{d}{dx}(\ln (1-x^2)) &= -\frac{2x}{1-x^2} \\
-\frac{2x}{1-x^2} &= \sum_{n=0}^\infty -2x(x^2)^n \\
&=-2\sum_{n=0}^\infty x^{2n+1} \\
\ln(1-x^2) &= \int -2\sum_{n=0}^\infty x^{2n+1} \mathrm{d}x \\
&=-2\sum_{n=0}^\infty \frac{x^{2n+2}}{2n+2} \\
&= -\sum_{n=0}^\infty \frac{x^{2n+2}}{n+1}
\end{align}
By expanding out the first few terms we can see that this is indeed the same as is confirmed above.
