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.

\begin{align*}I_n&=\int_0^1 \dfrac{x^n}{\sqrt{x^3+1}}dx=\int_0^1 \dfrac{x^{n-3}(x^3+1-1)}{\sqrt{x^3+1}}dx = \int_0^1 x^{n-3}\sqrt{x^3+1}dx - \int_0^1 \dfrac{x^{n-3}}{\sqrt{x^3+1}}dx\\ &= \int_0^1 x^{n-3}\sqrt{x^3+1}dx-I_{n-3}\end{align*}This integral is handled with integration by parts:
$$\int_0^1 x^{n-3}\sqrt{x^3+1}dx=\frac{x^{n-2}}{n-2}\sqrt{x^3+1}|_0^1-\frac{3}{2(n-2)}\int_0^1 \dfrac{x^{n}}{\sqrt{x^3+1}}dx=\dfrac{\sqrt{2}}{n-2}-\frac{3}{2(n-2)}I_n$$Therefore $$I_n=\dfrac{\sqrt{2}}{n-2}-\frac{3}{2(n-2)}I_n-I_{n-3}$$ or $$(n-2+\frac{3}{2})I_n+(n-2)I_{n-3}=\sqrt{2}\tag{1}$$Which is equivalent to your formula, so I would guess you're correct, though I haven't computed $I_2$ to check if $I_8$ is correct.
