Finding a closed form for the following integral: $\int_0^1\sqrt{1+x^k}dx$ Is there a nice closed form, for the following integral:
$$\int_0^1\sqrt{1+x^k}dx$$
And how can I derive it?
I have no idea how to get started, thanks for any help. This problem came up when I was working on a number theory problem in my university for some values of $k$.
 A: If we expand $\sqrt{1+x^k}$ in series and then intergrate  then we get a close representation in terms of hypergeometric function
$$
\int_0^1 \sqrt{1+x^k}={}_2F_1\left(-\frac{1}{2},\frac{1}{k};\frac{k+1}{k};-1 \right)
$$
A: Well, using IBP you can show that:
$$\mathscr{S}_\text{n}:=\int_0^1\sqrt{1+x^\text{n}}\space\text{d}x=\sqrt{2}-\frac{\text{n}}{2}\int_0^1\frac{x^\text{n}}{\sqrt{1+x^\text{n}}}\space\text{d}x\tag1$$
Using the fact that:
$$\frac{x^\text{n}}{\sqrt{1+x^\text{n}}}=\sqrt{1+x^\text{n}}-\frac{1}{\sqrt{1+x^\text{n}}}\tag2$$
We can write:
$$\mathscr{S}_\text{n}=\sqrt{2}-\frac{\text{n}}{2}\cdot\left\{\underbrace{\int_0^1\sqrt{1+x^\text{n}}\space\text{d}x}_{=\space\mathscr{S}_\text{n}}-\int_0^1\frac{1}{\sqrt{1+x^\text{n}}}\space\text{d}x\right\}\tag3$$
Using the Binomial series we know that we can write:
$$\mathscr{S}_\text{n}=\sqrt{2}-\frac{\text{n}}{2}\cdot\left\{\mathscr{S}_\text{n}-\int_0^1\sum_{\text{k}\ge0}\frac{\left(-1\right)^\text{k}\left(2\text{k}\right)!}{2^{2\text{k}}\left(\text{k}!\right)^2}\cdot x^{\text{nk}}\space\text{d}x\right\}=$$
$$\sqrt{2}-\frac{\text{n}}{2}\cdot\left\{\mathscr{S}_\text{n}-\sum_{\text{k}\ge0}\frac{\left(-1\right)^\text{k}\left(2\text{k}\right)!}{2^{2\text{k}}\left(\text{k}!\right)^2}\int_0^1x^{\text{nk}}\space\text{d}x\right\}=$$
$$\sqrt{2}-\frac{\text{n}}{2}\cdot\left\{\mathscr{S}_\text{n}-\underbrace{\sum_{\text{k}\ge0}\frac{\left(-1\right)^\text{k}\left(2\text{k}\right)!}{2^{2\text{k}}\left(\text{k}!\right)^2}\cdot\frac{1}{1+\text{nk}}}_{:=\space\mathcal{I}_\text{n}}\right\}\tag4$$
Solving for $\mathscr{S}_\text{n}$, gives:
$$\mathscr{S}_\text{n}=\frac{2\sqrt{2}+\text{n}\cdot\mathcal{I}_\text{n}}{2+\text{n}}\tag5$$
