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?)

$$\frac{d}{dx} \ln \left(\frac{1+x}{1-x} \right) = \frac{d}{dx} [\ln (1+x) -\ln (1-x)] = \frac{1}{1+x} + \frac{1}{1-x} = \frac{2}{(1+x)(1-x)} = \frac{2}{1-x^2}.$$
$$\ln \left(\frac{1+x}{1-x} \right) = 2\int \frac{dx}{1-x^2} = 2\int dx [1+x^2 +x^4 + x^6 + \cdots] = 2[x+x^3/3+x^5/5+x^7/7 \cdots]. $$
