1. ## Integration problem

Prove that if $I_n=\displaystyle \int x^n(1+x^3)^7 dx$ then
$I_n =\frac{1}{n+22}\left(x^{n-2}(1+x^3)^8-(n-2)I_{n-3}\right)$
(Hint. Use $x^n\equiv x^{n-2}x^2$)
Hence or otherwise determine $\displaystyle \int x^5(1+x^3)^7 dx$

My problem is in the first part.
$I_n=\displaystyle\int x^{n-2} x^2(1+x^3)^7 dx$
Integrating by parts
$=\displaystyle \frac{1}{24}x^{n-2}(1+x^3)^8-\frac{n-2}{24}\int x^{n-3}(1+x^3)^8 dx$

I can't find where to find the $\frac{1}{n+22}$ part.
Thanks!

2. Originally Posted by arze
Prove that if $I_n=\displaystyle \int x^n(1+x^3)^7 dx$ then
$I_n =\frac{1}{n+22}\left(x^{n-2}(1+x^3)^8-(n-2)I_{n-3}\right)$
(Hint. Use $x^n\equiv x^{n-2}x^2$)
Hence or otherwise determine $\displaystyle \int x^5(1+x^3)^7 dx$

My problem is in the first part.
$I_n=\displaystyle\int x^n x^2(1+x^3)^7 dx$
Integrating by parts
$=\displaystyle \frac{1}{24}x^{n-2}(1+x^3)^8-\frac{n-2}{24}\intx^{n-3}(1+x^3)^8 dx$

I can't find where to find the $\frac{1}{n+22}$ part.
Thanks!
After a correct application of integration by parts you get:

$\displaystyle I_n = \frac{x^{n-2}}{24} (1 + x^3)^8 - \frac{n-2}{24} \int x^{n-3} (1 + x^3)^8 \, dx$ .... (A)

Now note that

$\displaystyle \int x^{n-3} (1 + x^3)^8 \, dx = \int x^{n-3} (1 + x^3) (1 + x^3)^7 \, dx = \int x^{n-3} (1 + x^3)^7 \, dx + \int x^n (1 + x^3)^7 \, dx = I_{n-3} + I_n$ .... (B)

Substitute (B) into (A) and re-arrange to make $I_n$ the subject.