Finding the series representation of $\ln\left(\frac{1+x}{1-x}\right)$ 
Given that $\frac{1}{1-x}=\sum^{\infty}_{n=0}x^n$, what is the series representation of $\ln\left(\frac{1+x}{1-x}\right)$?  

Differentiating $\ln\left(\frac{1+x}{1-x}\right)$ results in: $\frac2{(1-x)^2}$. This means that $\ln\left(\frac{1+x}{1-x}\right) = 2\int\frac1{(1-x)^2}\ dx$
Can I now say that $\ln\left(\frac{1+x}{1-x}\right)=2\left(\sum^{\infty}_{n=0}\int x^n\right)^2$  ? This would get: $2\sum^{\infty}_{n=0}\frac{x^{2(n+1)}}{(n+1)^2}$.
I'm assuming the answer is wrong because the answer key did not agree with me. Did I mess up somewhere in this problem?     
The answer key only doubles the $x$ in this step: $\ln\left(\frac{1+x}{1-x}\right) = 2\int\frac1{(1-x)^2}\ dx$ so instead of $2\int\frac1{(1-x)^2}\ dx$ they get just $2\int\frac1{1-x^2}\ dx$. Why is this correct? (or is it incorrect?)

Differentiate:
$$f'(x)=\left ( \ln \frac {1+x}{1-x} \right)'=\frac {1-x}{1+x} \frac {2}{(1-x)^2} $$
$$f'(x)=\frac 2 {1-x^2}$$
$$f'(x)= 2\sum_{n=0}^\infty x^{2n}$$
Integrate:
$$\boxed {f(x)=2\sum_{n=0}^\infty \dfrac  { x^{2n+1}}{2n+1}}$$
