Question

Prove that ∫x^m(logx)ⁿdx for x ∈ [0, 1] = ((-1)ⁿ.n!)/(m + 1)^{n + 1}, where n is a positive integer and m > –1.

Answer 1

Substitute log x = – t or x = e– t

Then dx = – e^–t dt

when x = 0, t = ∞ and when x = 1, t = 0.

Therefore,