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If $|x|<1,$ Then the sum of the series $$ \frac{2x-1}{1-x+x^2}+\frac{4x^3-2x}{1-x^2+x^4}+\frac{8x^7-4x^3}{1-x^4+x^8}+\cdots \cdots $$

Try: Let $$A= \frac{1-2x}{1-x+x^2}+\frac{4x^3-2x}{1-x^2+x^4}+\frac{8x^7-4x^3}{1-x^4+x^8}+\cdots $$

$\displaystyle \int Adx $

$$= \int \bigg[\frac{1-2x}{1-x+x^2}+\frac{4x^3-2x}{1-x^2+x^4}+\frac{8x^7-4x^3}{1-x^4+x^8}+\cdots \cdots \bigg]dx$$

$$\int Adx = \ln\bigg[(1-x+x^2)\cdot (1-x^2+x^4)\cdot (1-x^4+x^8)\cdots \bigg]$$

Now i have seems that expression under $\ln$ on

Right side must have closed form in $-1<x<1$

But i could not understand how can i find it.

could some help me, thanks

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    $\begingroup$ Is the first term supposed to be $$\frac{2x-1}{1-x+x^2}\,?$$ Otherwise, your result is slightly wrong. (In the logarithm, you should have $\dfrac1{1-x+x^2}$ instead of $1-x+x^2$.) $\endgroup$ – Batominovski Oct 9 '18 at 7:32
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Hint:

$$(1+x+x^2)(1-x+x^2)=(1+x^2)^2-x^2=?$$

$$\implies\prod_{r=1}^n(1-x^{2^r}+\left(x^{2^r}\right)^2)=\dfrac{1-x^{2^{n+1}}+\left(x^{2^{n+1}}\right)^2}{1+x+x^2}$$

Now $\lim_{n\to\infty}2^{n+1}\to\infty$

and $|x|<1, \lim_{m\to\infty}x^m=0$

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Hint: $$(1+x+x^2)\bigg[(1-x+x^2)\cdot (1-x^2+x^4)\cdot (1-x^4+x^8)\cdots (1-x^{2^n}+x^{2^{n+1}})\bigg]=1+x^{2^{n+2}}$$

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