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!

$$\frac{1}{\sqrt{1-x}}=\sum_{n\geq 0}\frac{1}{4^n}\binom{2n}{n}x^n\tag{1} $$
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 $\frac{\Gamma^2\left(\frac{1}{4}\right)}{12\sqrt{\pi}}=\frac{\pi}{3\,\text{AGM}(\sqrt{2},2)}$.
