<s>[INST] Finding the value of an infinite sum from a definite integral: It can be computed that $$\int_0^1 \frac{x^2+1}{x^4+x^2+1}dx = \frac{\pi}{2\sqrt{3}}.$$ 
The integral can be done using $$\frac{2(x^2+1)}{x^4+x^2+1} = \frac{1}{x^2-x+1}+\frac{1}{x^2+x+1}.$$
But I got stuck in the following part:
We need to show that $$1-\frac{1}{5}+\frac{1}{7} - \frac{1}{11}+\frac{1}{13}- \cdots = \frac{\pi}{2\sqrt{3}}$$ using the above definite integral. The sum of the above series can be computed using Fourier series. Life becomes hard for me when the above sum is to be shown using the above integral. Any suggestions of help is much appreciated.
 [/INST] $$\frac{x^2+1}{x^4+x^2+1}
=\frac{(1-x^2)(x^2+1)}{(1-x^2)(x^4+x^2+1)}
=\frac{1-x^4}{1-x^6}=1-x^4+x^6-x^{10}+x^{12}-x^{16}+\cdots.$$
Now integrate this termwise.
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