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.

Integrating by parts, we have 
$$\begin{align}
I_n&=\int_0^1\frac{x^n}{(1+x^3)^{1/2}}dx\\\\
&=\left.\frac23 x^{n-2}(x^3+1)\right|_0^1-\frac23(n-2)\int_0^1x^{n-3}(x^3+1)^{1/2}dx \tag 1\\\\
&=\frac232^{1/2}-\frac23(n-2)\int_0^1\frac{x^{n-3}(x^3+1)}{(x^3+1)^{1/2}}\,dx \\\\
&=\frac232^{1/2}-\frac23 (n-2)I_n-\frac23(n-2)I_{n-3}\\\\
(2n-1)I_n+2(n-2)I_{n-3}&=2^{3/2} \tag 2\\\\
\end{align}$$
Using $(1)$, we can see that $I_2$ given by
$$I_2=\frac23(\sqrt{2}-1) \tag 3$$
Then, using $(2)$ and $(3)$ and iterating once to $I_5$ 
$$I_5=\frac29 (2-\sqrt{2})$$
and again to $I_8$ reveals 
$$I_8=\frac{2}{45}(7\sqrt{2}-8)$$
