CALCULUS-0-1-x-2n-1-x-1-dx-n-1-

Question Number 163709 by amin96 last updated on 09/Jan/22

CALCULUS

\int_{0}^{1} \frac{x^{2 n - 1}}{x + 1} dx = ? n ⩾ 1

Answered by Mathspace last updated on 09/Jan/22

\int_{0}^{1} \frac{x^{2 n - 1}}{1 + x} d x = \int_{0}^{1} x^{2 n - 1} \sum_{p = 0}^{\infty} {(- 1)}^{p} x^{p} d x

= \sum_{p = 0}^{\infty} {(- 1)}^{p} \int_{0}^{1} x^{p + 2 n - 1} d x

= \sum_{p = 0}^{\infty} \frac{{(- 1)}^{p}}{p + 2 n}