Can the Summation Expression Be Simplified Without Expanding Series?

[Dethrone]

I encountered this expression while trying o express $f(x)=\ln\left({\frac{1+x}{1-x}}\right)$ in terms of a power series:

$$\int \left[ \sum_{n=0}^{\infty}(-1)^n x^n + \sum_{n=0}^{\infty}x^n\right] \,dx$$
The book simplifies this expression as $\int \sum_{n=0}^{\infty}2x^{2n} \,dx$ by expanding the above two series and simplifying. But, is there a way to simplify the expression algebraically without expanding any series?Here's my attempt:
$$\int \sum_{n=0}^{\infty} \left[(-1)^nx^n+x^n\right] \,dx$$
$$\int \sum_{n=0}^{\infty} x^n\left[\left(-1)^n+1\right)\right] \,dx$$

I can't seem to get here:
$$\int \sum_{n=0}^{\infty}2x^{2n} \,dx$$

Edit: I see that $\int \sum_{n=0}^{\infty} x^n\left[\left(-1)^n+1\right)\right] \,dx$ in my attempt above contains only even values of $n$, since the odd values are 0. Could the key step be realizing that and multiplying $n$ by two?

 

[MarkFL]

All of the terms where $n$ is odd are zero, so you only need consider when $n$ is even. So, that's why $2n$ is used. :D

 

[Prove It]

I encountered this expression while trying o express $f(x)=\ln\left({\frac{1+x}{1-x}}\right)$ in terms of a power series:

$$\int \left[ \sum_{n=0}^{\infty}(-1)^n x^n + \sum_{n=0}^{\infty}x^n\right] \,dx$$
The book simplifies this expression as $\int \sum_{n=0}^{\infty}2x^{2n} \,dx$ by expanding the above two series and simplifying. But, is there a way to simplify the expression algebraically without expanding any series?Here's my attempt:
$$\int \sum_{n=0}^{\infty} \left[(-1)^nx^n+x^n\right] \,dx$$
$$\int \sum_{n=0}^{\infty} x^n\left[\left(-1)^n+1\right)\right] \,dx$$

I can't seem to get here:
$$\int \sum_{n=0}^{\infty}2x^{2n} \,dx$$

Edit: I see that $\int \sum_{n=0}^{\infty} x^n\left[\left(-1)^n+1\right)\right] \,dx$ in my attempt above contains only even values of $n$, since the odd values are 0. Could the key step be realizing that and multiplying $n$ by two?

$\displaystyle \begin{align*} \ln{ \left( \frac{1 + x}{1 - x} \right) } \equiv \ln{ \left( 1 + x \right) } - \ln{ \left( 1 - x \right) } \end{align*}$

Now notice $\displaystyle \begin{align*} \frac{\mathrm{d}}{\mathrm{d}x } \left[ \ln{ \left( 1 + x \right) } \right] = \frac{1}{1 + x} \end{align*}$ and $\displaystyle \begin{align*} \frac{\mathrm{d}}{\mathrm{d}x} \left[ \ln{ \left( 1 - x \right) } \right] = -\frac{1}{1 - x} \end{align*}$. Each of those expressions is the closed form of a geometric series...

$\displaystyle \begin{align*} \frac{1}{1 + x} &= \frac{1}{1 - \left( -x \right) } \\ &= \sum_{n = 0}^{\infty}{ \left( -x \right) ^n } \textrm{ for } \left| - x \right| < 1 \\ &= \sum_{n = 0}^{\infty} { \left( -1 \right) ^n \, x^n } \textrm{ for } \left| x \right| < 1 \end{align*}$

so

$\displaystyle \begin{align*} \ln{ \left( 1 + x \right) } &= \int{ \sum_{n = 0}^{\infty}{\left( -1 \right) ^n \, x^n } \,\mathrm{d}x} \\ &= \sum_{n = 0}^{\infty}{ \frac{ \left( -1 \right) ^n \, x^{n + 1}}{n + 1} } + C \textrm{ also for } \left| x \right| < 1 \end{align*}$

and because we know $\displaystyle \begin{align*} \ln{(1)} = 0 \end{align*}$, when we let $\displaystyle \begin{align*} x = 0 \end{align*}$ we have

$\displaystyle \begin{align*} \ln{ \left( 1 + 0 \right) } &= \sum_{n =0}^{\infty} { \frac{\left( -1 \right) ^n \, 0^{n + 1}}{n + 1} } + C \\ \ln{(1)} &= 0 + C \\ 0 &= C \end{align*}$

and thus $\displaystyle \begin{align*} \ln{ \left( 1 + x \right) } = \sum_{n = 0}^{\infty}{ \frac{ \left( -1 \right) ^n \, x^{n + 1} }{n + 1} } \end{align*}$ which is convergent for $\displaystyle \begin{align*} \left| x \right| < 1 \end{align*}$.Now follow a similar process to get a series for $\displaystyle \begin{align*} -\frac{1}{1 - x} \end{align*}$, integrate it to get a series for $\displaystyle \begin{align*} \ln{ \left( 1 - x \right) } \end{align*}$, and then see what you get for $\displaystyle \begin{align*} \ln{ \left( 1 + x \right) } - \ln{ \left( 1 - x \right) } \end{align*}$.