<s>[INST] Logarithmic formula By making as few assumptions as possible, prove that 
$$\log \left({{1+x} \over {1-x}} \right)=2 \left(\frac{x}{1}+\frac{x^3}{3}+\frac{x^5}{5}+\cdots\right)$$
for $|x|<1$.
 [/INST] Note that for $|x|<1$, we can write
$$\begin{align}
\log\left(\frac{1+x}{1-x}\right)&=\int_0^{x}\color{#034da3}{\left(\frac{1}{1+u}+\frac{1}{1-u}\right)}\,du\\\\
&=\int_0^x \color{#034da3}{\frac{2}{1-u^2}}\,du\\\\
&=\int_0^x \overbrace{\color{#034da3}{2\sum_{n=0}^\infty u^{2n}}}^{\color{#034da3}{\text{Geometric Series}}}\,du\\\\
&=2\sum_{n=0}^\infty \frac{x^{2n+1}}{2n+1}\\\\
&=2\sum_{n=1}^\infty\frac{x^{2n-1}}{2n-1}
\end{align}$$

Alternatively, for $|x|<1$, we can write
$$\begin{align}
\log\left(\frac{1+x}{1-x}\right)&=\int_0^{x}\color{#034da3}{\left(\frac{1}{1+u}+\frac{1}{1-u}\right)}\,du\\\\
&=\int_0^x \overbrace{\color{#034da3}{\sum_{n=0}^\infty (1+(-1)^n)u^n}}^{\color{#034da3}{\text{Geometric Series of the integrand}}}\,dx\\\\
&=\underbrace{\sum_{n=0}^\infty\frac{(1+(-1)^n)x^{n+1}}{n+1}}_{\text{After integrating the power series term-by-term}}\\\\
&=2\sum_{n=0}^\infty \frac{x^{2n+1}}{2n+1}\\\\
\end{align}$$
</s>