How to find the power series of $\sqrt{1+x^4}$? The complete question is to find the integral from $0$ to $1$ of $$\sqrt{1+x^4}$$
I am unsure of how to find the power series of this equation in order to do that. I haven't dealt with square root power series equations yet and any help would be appreciated. Thank you!
 A: $$\frac{1}{\sqrt{1-x}}=\sum_{n\geq 0}\frac{1}{4^n}\binom{2n}{n}x^n\tag{1} $$
for any $x\in[-1,1)$ is a standard result. By applying termwise integration one gets
$$ \sqrt{1-x} = \sum_{n\geq 0}\frac{1}{4^n}\binom{2n}{n}\frac{x^n}{1-2n}\tag{2} $$
for any $x\in[-1,1]$. By replacing $x$ with $-x^4$ one gets
$$ \sqrt{1+x^4} = \sum_{n\geq 0}\frac{(-1)^n}{4^n}\binom{2n}{n}\frac{x^{4n}}{1-2n}\tag{3}$$
and finally
$$ \int_{0}^{1}\sqrt{1+x^4}\,dx=\sum_{n\geq 0}\frac{(-1)^n}{4^n}\binom{2n}{n}\frac{1}{(1-2n)(4n+1)}.\tag{4} $$
By partial fraction decomposition, the RHS of $(4)$ only depends on $\sqrt{2}$ and
$$ \sum_{n\geq 0}\frac{1}{4^n}\binom{2n}{n}\frac{(-1)^n}{(4n+1)}=\frac{\Gamma^2\left(\tfrac{1}{4}\right)}{8\sqrt{\pi}}.\tag{5}$$
The last equality can be derived through Euler's Beta function. Summarizing,
$$ \int_{0}^{1}\sqrt{1+x^4}\,dx = \phantom{}_2 F_1\left(-\tfrac{1}{2},\tfrac{1}{4};\tfrac{5}{4};-1\right)= \frac{\sqrt{2}}{3}+\frac{\Gamma^2\left(\frac{1}{4}\right)}{12\sqrt{\pi}}.\tag{6}$$
A numerical computation is extremely simple via $\frac{\Gamma^2\left(\frac{1}{4}\right)}{12\sqrt{\pi}}=\frac{\pi}{3\,\text{AGM}(\sqrt{2},2)}$.
A: In order to find the power series of $\sqrt{1+x}$, you use the binomial series:$$\sqrt{1+x}=(1+x)^{\frac12}=\sum_{n=0}^\infty\binom{\frac12}nx^n$$and therefore$$\sqrt{1+x^4}=\sum_{n=0}^\infty\binom{\frac12}nx^{4n}.$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
\int_{0}^{1}\root{1 + x^{4}}\dd x &
\,\,\,\stackrel{x^{4}\ \mapsto\ x}{=}\,\,\,
{1 \over 4}\int_{0}^{1}x^{-3/4}\,\pars{1 + x}^{1/2}\,\dd x
\\[5mm] & =
{1 \over 4}\int_{0}^{1}x^{-3/4}\,\pars{1 - x}^{0}\,
\bracks{\vphantom{\Large A}1 - \pars{-1}x}^{1/2}\,\dd x
\end{align}
The integral is an
Euler Type Hypergeometric Function. The above expression is reduced to
\begin{align}
&\overbrace{{1 \over 4}\bracks{\mrm{B}\pars{{1 \over 4},1}}}^{\ds{=\ 1}}\,\,\,\,
\mbox{}_{2}\mrm{F}_{1}\pars{-\,{1 \over 2};{1 \over 4};
{5 \over 4},-1}
\\[5mm] = &\
\bbx{\mbox{}_{2}\mrm{F}_{1}\pars{-\,{1 \over 2};{1 \over 4};
{5 \over 4},-1}} \approx 1.0894
\end{align}

$\ds{\mrm{B}}$ is the
  Beta Function.

