I'll appreciate some help with this one:
Hint: let
$\displaystyle \displaystyle g'(x)=\frac{1}{1-x^4}=\sum_{n=0}^{\infty}x^{4n}$
Now integrate term by term to get
$\displaystyle \displaystyle g(x)=\sum_{n=0}^{\infty}\frac{x^{4n+1}}{4n+1}=\int \frac{1}{1-x^4}dx$
Use partial fractions and integrate and let $\displaystyle x=\frac{1}{2}$
Is...
$\displaystyle \displaystyle f^{'} (x) = 1+x^{4}+x^{8} + ... + x^{4 n} + ... = \frac{1}{1-x^{4}} = \frac{\frac{1}{2}}{1+x^{2}} + \frac{\frac{1}{2}}{1-x^{2}}$ (1)
... so that is...
$\displaystyle \displaystyle f(x) = \int_{0}^{x} \frac{dt}{1-t^{4}} = \frac{1}{2}\ \tan^{-1} x + \frac{1}{4}\ \ln \frac{1+x}{1-x}$ (2)
... and from (2)...
$\displaystyle \displaystyle \sum_{n=0}^{\infty} \frac{1}{2^{4 n}\ (4n+1)} = 2\ f(\frac{1}{2}) = \tan^{-1} \frac{1}{2} + \ln \sqrt{3}$ (3)
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$