Prove that $\int(x^2-1)^n \,dx = \frac{x(x^2-1)^n}{2n+1} - \frac{2n}{2n+1}\int(x^2-1)^{n-1} \,dx$

Question

I have tried to solve the problem mainly with the LS of the equation. I can not seem to get rid of the x variable within the resultant integrand. ex. after the first integration by parts I am left with: $x(x^2-1)^n - 2n\int x^2(x^2-1)^{n-1} \,dx$

Thanks for all the help in advance!

Answer 1

Hint: Just write $\int x^{2}(x^{2}-1)^{n-1}dx$ as $\int (x^{2}-1+1)(x^{2}-1)^{n-1}dx=\int (x^{2}-1)^{n}dx+\int (x^{2}-1)^{n-1}dx$ and transfer one term to the left side.