Finding the series representation of $\ln\left(\frac{1+x}{1-x}\right)$

Question

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?)

Answer 1

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

Answer 2

Integrate the given equation \begin{eqnarray*} \ln\left( \frac{1}{1-x}\right) =\sum_{n=0}^{\infty} \frac{x^{n+1}}{n+1}. \\ \end{eqnarray*} Now (with the given equation, again) sub $ x \rightarrow -x$ and integrate \begin{eqnarray*} \ln( 1+x) =\sum_{n=0}^{\infty} \frac{(-1)^n x^{n+1}}{n+1}. \\ \end{eqnarray*} Add these and note that the even powers cancel, thus we have \begin{eqnarray*} \ln\left( \frac{1+x}{1-x}\right) = 2 \sum_{n=0}^{\infty} \frac{x^{2n+1}}{2n+1}. \\ \end{eqnarray*}

Answer 3

$$\frac{d}{dx} \ln \left(\frac{1+x}{1-x} \right) = \frac{d}{dx} [\ln (1+x) -\ln (1-x)] = \frac{1}{1+x} + \frac{1}{1-x} = \frac{2}{(1+x)(1-x)} = \frac{2}{1-x^2}.$$

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

Answer 4

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: $$\boxed {f(x)=2\sum_{n=0}^\infty \dfrac { x^{2n+1}}{2n+1}}$$