<s>[INST] 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?
 [/INST] 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 $J_n=2\sqrt2$.
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