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.

Perform the integration in steps:

$$
\begin{array}{rcl}
I_8
&=& \displaystyle \int_0^1 \frac{x^8}{\sqrt{x^3 + 1}} dx
    = \displaystyle \int_0^1 x^6 \frac{x^2}{\sqrt{x^3 + 1}} dx\\
&=& \displaystyle
    \left[ \frac{2}{3} x^6 \sqrt{x^3+1} \right]_0^1
        - \int_0^1 4 x^3 x^2 \sqrt{x^3+1} dx\\
&=& \displaystyle
    \left[ \frac{2}{3} x^6 \sqrt{x^3+1} \right]_0^1
    - \left[ \frac{8}{9} x^3 \sqrt{x^3+1}^3 \right]_0^1
        + \int_0^1 \frac{8}{3} x^2 \sqrt{x^3+1}^3 dx\\
&=& \displaystyle
    \left[ \frac{2}{3} x^6 \sqrt{x^3+1} \right]_0^1
    - \left[ \frac{8}{9} x^3 \sqrt{x^3+1}^3 \right]_0^1
        +  \left[ \frac{16}{45} \sqrt{x^3+1}^5 \right]_0^1\\
&=& \displaystyle
    \left[ \frac{2}{3} x^6 \sqrt{x^3+1}
    - \frac{8}{9} x^3 \sqrt{x^3+1}^3
    + \frac{16}{45} \sqrt{x^3+1}^5 \right]_0^1\\
&=& \displaystyle
    \left( \frac{2}{3} - \frac{16}{9} + \frac{64}{45} \right) \sqrt{2} - \frac{16}{45}\\
&=& \displaystyle \bbox[16px,border:2px solid #800000] {\frac{14}{45} \sqrt{2} - \frac{16}{45}}
\end{array}
$$

