Given $I_n = \int_0^1 \frac{x^n}{\sqrt{x^3+1}} dx$, show $(2n-1)I_n+2(n-2)I_{n-3}=2\sqrt{2}$

Question

Let $I_n = \int_0^1 \frac{x^n}{\sqrt{x^3+1}} dx$. Show that $(2n-1)I_n+2(n-2)I_{n-3}=2\sqrt{2}$ for all $n \geq 3$.

Tried using integration by part straight away, but it doesn't work.

Note: $I_2$ can be easily calculated by using substitution, and we also know that $I_5$ can be reduced to $I_2$, $I_8$ can be reduced to $I_5$, and so on, but $I_2$ is out of the requirement $n \geq 3$.

Answer 1

HINT:

Integrate by parts $u = x^{n-2}$ and $dv = \frac{x^2}{\sqrt{x^3 + 1}} \ dx$.

Answer 2

Integrate by parts

$$\begin{align} I_n &= \int_0^1 \frac{x^n}{\sqrt{x^3+1}} dx\\ & =\frac23 \int_0^1 x^{n-2}d(\sqrt{x^3+1} )dx\\ & =\frac23 \sqrt2-\frac23(n-2) \int_0^1 x^{n-3}\sqrt{x^3+1} dx\\ & =\frac23 \sqrt2-\frac23(n-2) \int_0^1 \frac{x^{n-3}(x^3+1)}{\sqrt{x^3+1}} dx\\ & =\frac23 \sqrt2-\frac23(n-2) (I_n+I_{n-3})\\ \end{align}$$

Thus, $$(2n-1)I_n+2(n-2)I_{n-3}=2\sqrt{2}$$