Edwards Calculus chapter 4 problem 81 I can see that the terms arise from the differentiation of $\log(1+x^{2^r} )$ and tried to do it but couldn't figure it out in the end.

Show that
  $$
\frac{1}{1+x} + \frac{2x}{1+x^2} + \frac{4x^3}{1+x^4} + \frac{8x^7}{1+x^8} + \ldots = \sum_{n=0}^{\infty} \frac{2^{n} x^{(2^n -1)}}{1+x^{(2^n)}} =  \frac{1}{1-x} \qquad \text{when}~|x|<1
$$

 A: Let's assume 
$$S=\dfrac{1}{1+x}+\dfrac{2x}{1+x^2}+\dfrac{4x^3}{1+x^4}+...$$
The common term of the series, as seen from the pattern, is $\displaystyle \frac{2^{n} x^{(2^n -1)}}{1+x^{(2^n)}}$ for $n=0,1,2,3,...$ and the sum is then $\displaystyle\sum_{n=0}^{\infty} \frac{2^{n} x^{(2^n -1)}}{1+x^{(2^n)}}$.
Now, noting that the numerator in each case is the derivative of the denominator, as you've already done, we integrate $$\int_0^x Sdx=\left[\ln(1+x)+\ln(1+x^2)+\ln(1+x^4)+...\right]^x_0$$
$$=\ln[(1+x)(1+x^2)(1+x^4)...]$$
$$=\ln(1+x+x^2+x^3...)$$
Here, $\ln$ denotes the natural logarithm and I've used the facts that for the lower limit, $\ln1=0$, then $\ln a+\ln b= \ln{ab}$ and expanded the multiplication in the last step.
Note that the series $1+x+x^2+x^3+...$ is the series expansion of $\dfrac{1}{1-x}$ and this is valid only when $|x|<1$. So the term inside the logarithm converges provided that $|x|<1$ and the sum is $\dfrac{1}{1-x}$. 
Thus, $\displaystyle\int_0^x Sdx=\ln {\dfrac{1}{1-x}}$ and so, 
$$ S=\dfrac{d}{dx} \ln\left( \dfrac{1}{1-x} \right)=\dfrac{1}{1-x}$$
