Edwards Differential Calculus for Beginners, 1896. Chapter 4, Question 82 This question asks for an evaluation of an infinite series assuming only knowledge of basic differential calculus. I couldn't figure it out, but user 'Dr. MV' gave me a hint which was sufficient for me to find the answer. I hope this makes the question seem more relevant, I'm new around these parts and have yet to learn what makes for adequate structure in regards to asking questions.
Here is the question.
Prove that if $x <1$, then
$$\sum_{k=0}^\infty\frac{2^kx^{2^k-1}-2^{k+1}x^{2^{k+1}-1}}{1-x^{2^k}+x^{2^{k+1}}}=\frac{1-2x}{1-x+x^2} + \frac{2x-4x^3}{1-x^2+x^4} + \frac{4x^3-8x^7}{1-x^4+x^8} + ... =\frac{1+2x}{1+x+x^2}$$

HINT:
$$\frac{2^kx^{2^k-1}-2^{k+1}x^{2^{k+1}-1}}{1-x^{2^k}+x^{2{k+1}}} =-\frac{d}{dx}\log\left(1-x^{2^k}+x^{2^{k+1}}\right)\tag 1$$
Then, write the argument of the logarithm in $(1)$ as 
$$1-x^{2^k}+x^{2^{k+1}}=\frac{1+x^{2^{k+1}}+x^{2^{k+2}}}{1+x^{2^k}+x^{2^{k+1}}}$$
Then, telescope and differentiate.
