Obtaining a series expansion for an integral I am trying to obtain a series expansion for the integral 
$\int_0^{1/2} dx/(1+x^4)$ 
So far I used the fact for a Geometric series $\sum_{k=0}^\infty r^k = \frac{1}{1-r}$ if $\lvert{r}\rvert<1$
and replaced $r$ with $-x^4$ and got $\frac{1}{1+x^4} = \sum_{k=1}^\infty (-1)^k x^{4k}$ if $\lvert{x}\rvert<1$
I am kind of unsure if this is even an answer or that if I need to go down a different path in order to solve this problem. If someone could tell me what to do in addition to this or what to differently or just show me how to do these types of problems it would be very appreciated. Thank you.
 A: Yes, your approach us fine.
Integrating term-by-term, your integral equals
$$\sum_{n\ge\color{red}{0}}\frac{(-1)^k}{4k+1}\frac1{2^{4k+1}}$$
I doubt if a nice closed form exists...
A: Incomplete Beta Function
$$
\begin{align}
\int_0^{1/2}\frac{\mathrm{d}x}{1+x^4}
&=\frac14\int_0^{1/16}\frac{x^{-3/4}\,\mathrm{d}x}{1+x}\\
&=\bbox[5px,border:2px solid #C0A000]{\operatorname{B}\left(\frac1{17};\frac14,\frac34\right)}
\end{align}
$$
Partial Fractions with $\alpha^2=i$; i.e. $\alpha=\frac{1+i}{\sqrt2}$ and $\alpha^3=\frac{-1+i}{\sqrt2}$
$$
\begin{align}
\int_0^{1/2}\frac{\mathrm{d}x}{1+x^4}
&=\int_0^{1/2}\frac12\left(\frac1{1-i^3x^2}+\frac1{1-ix^2}\right)\mathrm{d}x\\
&=\int_0^{1/2}\frac14\left(\frac1{1-\alpha^3x}+\frac1{1+\alpha^3x}+\frac1{1-\alpha x}+\frac1{1+\alpha x}\right)\mathrm{d}x\\
&=\frac1{4\alpha^3}\int_0^{\alpha^3/2}\left(\frac1{1-x}+\frac1{1+x}\right)\mathrm{d}x\\
&+\frac1{4\alpha}\int_0^{\alpha/2}\left(\frac1{1-x}+\frac1{1+x}\right)\mathrm{d}x\\
&=\frac1{4\alpha^3}\log\left(\frac{1+\alpha^3/2}{1-\alpha^3/2}\right)+\frac1{4\alpha}\log\left(\frac{1+\alpha/2}{1-\alpha/2}\right)\\
&=\frac{-1-i}{4\sqrt2}\log\left(\frac{2\sqrt2-1+i}{2\sqrt2+1-i}\right)+\frac{1-i}{4\sqrt2}\log\left(\frac{2\sqrt2+1+i}{2\sqrt2-1-i}\right)\\
&=\frac1{4\sqrt2}\log\left(\frac{5+2\sqrt2}{5-2\sqrt2}\right)
+\frac1{2\sqrt2}\arctan\left(\frac{2\sqrt2}3\right)\\
&=\bbox[5px,border:2px solid #C0A000]{\frac{\log\left(\frac{33+20\sqrt2}{17}\right)+\arctan\left(12\sqrt2\right)}{4\sqrt2}}
\end{align}
$$
