Integration of $I_n=\int_{0}^{1} \frac{x^n}{\sqrt{x^3+1}}dx$ Let $I_n=\int_{0}^{1} \frac{x^n}{\sqrt{x^3+1}}dx$. Show that $(2n−1)I_n+2(n−2)I_{n-3}= 2\sqrt{2}$ for all $n\geq3$. Then compute $I_8$.
How do I compute that integral? My idea was to reduce it to some known forms, as $\int\frac{1}{\sqrt{1-x^2}}dx$... Is there any way to solve the problem without computing the integral though?
 A: Just for the sake of amusement in solving integrals.
You can always mind in this way. The integration is bounded between $0$ and $1$ hence you may make use of the binomial series:
$$\frac{1}{\sqrt{x^3+1}} = (1+x^3)^{-1/2} = \sum_{k = 0}^{+\infty} \binom{-1/2}{k} x^{3k}$$
Hence
$$\int_0^1 \frac{x^n}{\sqrt{1+x^3}}\ \text{d}x = \sum_{k = 0}^{+\infty} \binom{-1/2}{k}\int_0^1 x^{3k + n}\ \text{d}x = \sum_{k = 0}^{+\infty} \binom{-1/2}{k} \frac{1}{3k + n + 1}$$
The last series does converge and its sum is expressed in terms of HyperGeometric Function:
$$\sum_{k = 0}^{+\infty} \binom{-1/2}{k} \frac{1}{3k + n + 1} = \frac{\, _2F_1\left(\frac{1}{2},\frac{n+1}{3};\frac{n+4}{3};-1\right)}{n+1}$$
A: I doubt that it is a good idea to use the hypergeometric function. Also substitution is not working. Actually it is pretty easy to use integration by parts to get the answer. Define 
$$ J_n=(2n−1)I_n+2(n−2)I_{n-3}, n\ge 3.$$
Note
\begin{eqnarray}
J_n&=&(2n−1)I_n+2(n−2)I_{n-3}\\
&=&\int_0^{1}\frac{(2n-1)x^n+(2n-4)x^{n-3}}{\sqrt{x^3+1}}dx\\
&=&\int_0^{1}\frac{x^{n-3}\bigg[(2n-1)x^3+(2n-4)\bigg]}{\sqrt{x^3+1}}dx\\
&=&\int_0^{1}\frac{x^{n-3}\bigg[(2n-1)(x^3+1)-3\bigg]}{\sqrt{x^3+1}}dx\\
&=&(2n-1)\int_0^{1}x^{n-3}\sqrt{x^3+1}dx-3\int_0^{1}\frac{x^{n-3}}{\sqrt{x^3+1}}dx\\
&=&\frac{2n-1}{n-2}\int_0^{1}\sqrt{x^3+1}d(x^{n-2})-3\int_0^{1}\frac{x^{n-3}}{\sqrt{x^3+1}}dx\\
&=&\frac{2n-1}{n-2}\left(\sqrt{x^3+1}x^{n-2}\bigg|_0^1-\frac{3}{2}\int_0^1\frac{x^{n}}{\sqrt{x^3+1}}dx\right)-3\int_0^{1}\frac{x^{n-3}}{\sqrt{x^3+1}}dx\\
&=&\frac{2n-1}{n-2}\sqrt2-\frac{3}{2n-4}J_n\\
\end{eqnarray}
and hence
$$ \left(1+\frac{3}{2n-4}\right)J_n=\frac{2n-1}{n-2}\sqrt2$$
which gives $J_n=2\sqrt2$.
A: HINT
Notice that if you let $u = x^3+1$ then $du = 3x^2dx$ so
$$
\int \frac{x^2 dx}{\sqrt{x^3+1}} = \frac{1}{3} \int u^{-1/2}du
$$
and apply integration by parts to your original integral
A:                     **This is another great way to compute $I_8$.**

As we have $I_8=\int_0^1\frac{x^8}{\sqrt{1+x^3}}dx$. Let $1+x^3=t^2$. You may ask me why am I doing this? I do it because we can do that by binomial_differentials. So you can overcome the indefinite integral first and the do for interval $[0,1]$.
A: This is a partial solution. There is a problem at the end. If you can resolve it, please
do so in a comment and I will fix my solution, or post it as a separate solution.
Here is a correct solution that avoids the problem of going from the incomplete beta function to Gauss's hypergeometric function.
Let $y=x^{3}$
\begin{align}
\int\limits_{0}^{1} \frac{x^{n}}{\sqrt{x^{3}+1}} dx 
&= \frac{1}{3} \int\limits_{0}^{1} \frac{y^{(n-2)/3}}{\sqrt{y+1}} dy \\
&= \frac{1}{3} \frac{\Gamma(\frac{n+1}{3})\Gamma(1)}{\Gamma(\frac{n+4}{3})}\,{}_{2}\mathrm{F}_{1}\left(\frac{1}{2},\frac{n+1}{3};\frac{n+4}{3};-1\right) \\
&= \frac{1}{n+1} \,{}_{2}\mathrm{F}_{1}\left(\frac{1}{2},\frac{n+1}{3};\frac{n+4}{3};-1\right)
\end{align}
We used the analytic continuation of Gauss's hypergeometric function
\begin{equation}
{}_{2}\mathrm{F}_{1}(a,b;c;z) 
= \frac{\Gamma(c)}{\Gamma(b)\Gamma(c - b)} \int\limits_{0}^{1} t^{b-1} (1-t)^{c-b-1} (1-zt)^{-a} dt
\end{equation}
For $\mathrm{Re}\, c \gt \mathrm{Re}\, b \gt 0$, $|\mathrm{arg}(1-z)| \lt \pi$
