<s>[INST] 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?)
 [/INST] Your derivative of the logarithm is wrong. It is not $\frac 2{(1-x)^2}$. It is $\frac{2}{1-x^2}$.
$$\left[\ln\frac{1+x}{1-x}\right]'=\frac{\left[\frac{1+x}{1-x}\right]'}{\frac{1+x}{1-x}}=\frac{\frac 2{(1-x)^2}}{\frac{1+x}{1-x}}=\frac{2}{(1+x)(1-x)}=\frac{2}{1-x^2}$$
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