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.

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.
