How can I find the reduction formula for $I_n=\int^1_0{x^6 (1+x^3)^{n}}dx$

Question

I need to find the reduction formula for $I_n=\int^1_0{x^6 (1+x^3)^{n}}dx$ so that I may find the relationship between $I_4$ and $I_3$, I have done the following:

Setting up for Integration by parts

$$u=(1+x^3)^n \implies u^{'}=3nx^2(1+x^3)^{n-1}$$ $$v^{'}=x^6 \implies v=\frac{1}{7}x^7$$ Applying Integration by parts

$$I_n=[\frac{1}{7}x^7(1+x^3)^n]^1_0 - \frac{3n}{7}\int^1_0{x^9(1+x^3)^{n-1}dx}$$ $$=\frac{2^n}{7}-\frac{3n}{7}\int^1_0{x^9(1+x^3)^{n-1}dx}$$

However I have no idea how to bring the integral of $\frac{3n}{7}\int^1_0{x^9(1+x^3)^{n-1}dx}$ into the form of $\int^1_0{x^6(1+x^3)^{n-1}dx}$ so that I may substitute it as $I_{n-1}$.

How can I do it, so that I may find a relationship between $I_4$ and $I_3$ ?

Answer 1

Notice that $$ I_n - I_{n - 1} = \int_0^1 {x^9 (1 + x^3 )^{n - 1} dx} . $$

Answer 2

For completeness, by $$x^{6}\left(1+x^{3}\right)^{n}-x^{6}\left(1+x^{3}\right)^{n-1}=x^{6}\left(1+x^{3}\right)^{n-1}\left(1+x^{3}-1\right)= x^{9}\left(1+x^{3}\right)^{n-1}, $$ we have $$ \begin{aligned} I_{n} &=\frac{2^{n}}{7}-\frac{3 n}{7}\left(I_{n}-I_{n-1}\right) \\ 7I_{n} &=2^{n}-3 n I_{n}+3 n I_{n-1} \\ I_{n} &=\frac{1}{7+3 n}\left(2^{n}+3 n I_{n-1}\right) \end{aligned} $$