Question

(x^2-1)^n在【0 1】求定积分

Answer 1

∫[0,1] (x^2-1)^n dx

=∫[0,1] (x-1)^n (x+1)^n d(x-1)

=1/(n+1) * ∫[0,1] (x+1)^n d(x-1)^(n+1)

=n/(n+1) * ∫[0,1] (x+1)^(n-1) (x-1)^(n+1) dx

=n(n-1)/(n+1)(n+2) * ∫[0,1] (x+1)^(n-2) (x-1)^(n+2) dx

=...

=n!* n!/(2n)!* ∫[0,1] (x-1)^(2n) dx

=n!* n!/(2n+1)!