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.
 A: Perform the integration in steps:

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

A: 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 
$$I_8=\frac{2}{45}(7\sqrt{2}-8)$$
A: \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.
A: 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 $x\mapsto(x-1)^{1/3}$:
$$
\begin{align}
I_8
&=\int_0^1\frac{x^8}{\sqrt{x^3+1}}\,\mathrm{d}x\\
&=\frac13\int_1^2\frac{(x-1)^2}{\sqrt{x}}\,\mathrm{d}x\\
&=\frac13\left[\frac25x^{5/2}-\frac43x^{3/2}+2x^{1/2}\right]_1^2\\
&=\bbox[5px,border:2px solid #C0A000]{\frac{14\sqrt2-16}{45}}\tag{5}
\end{align}
$$
