<s>[INST] Am I getting the right answer for the integral $I_n= \int_0^1 \frac{x^n}{\sqrt {x^3+1}}\, dx$? 
Let $I_n= \int_0^1 \dfrac{x^n}{\sqrt {x^3+1}}\, dx$. Show that $(2n-1)I_n+2(n-2)I_{n-3}=2 \sqrt 2$ for all $n \ge 3$. Then compute $I_8$.

I get an answer for $I_8={{2 \sqrt 2} \over 135}(25-16 \sqrt 2)$, could somebody please check against my answer and see if I made a mistake.
 [/INST] If we set
$$
I_n=\int_0^1\frac{x^n}{\sqrt{x^3+1}}\,\mathrm{d}x\tag{1}
$$
Then, integration by parts gives
$$
\begin{align}
I_{n+3}+I_n
&=\int_0^1\frac{x^n(x^3+1)}{\sqrt{x^3+1}}\,\mathrm{d}x\\
&=\int_0^1x^n\sqrt{x^3+1}\ \mathrm{d}x\\
&=\frac1{n+1}\int_0^1\sqrt{x^3+1}\ \mathrm{d}x^{n+1}\\
&=\frac1{n+1}\left[\sqrt2-\int_0^1\frac32\frac{x^{n+3}}{\sqrt{x^3+1}}\,\mathrm{d}x\right]\tag{2}
\end{align}
$$
Applying a bit of algebra to $(2)$ yields
$$
\left(2n+5\right)I_{n+3}+(2n+2)I_n=2\sqrt2\tag{3}
$$
which, after substituting $n\mapsto n-3$, gives
$$
\bbox[5px,border:2px solid #C0A000]{(2n-1)I_n+(2n-4)I_{n-3}=2\sqrt2}\tag{4}
$$
Next, substitute $x\mapsto(x-1)^{1/3}$:
$$
\begin{align}
I_8
&=\int_0^1\frac{x^8}{\sqrt{x^3+1}}\,\mathrm{d}x\\
&=\frac13\int_1^2\frac{(x-1)^2}{\sqrt{x}}\,\mathrm{d}x\\
&=\frac13\left[\frac25x^{5/2}-\frac43x^{3/2}+2x^{1/2}\right]_1^2\\
&=\bbox[5px,border:2px solid #C0A000]{\frac{14\sqrt2-16}{45}}\tag{5}
\end{align}
$$
</s>