let-U-n-0-1-x-n-1-x-dx-calculate-U-n-U-n-1-

Question Number 78261 by msup trace by abdo last updated on 15/Jan/20

l e t U_{n} = \int_{0}^{1} \frac{x^{n}}{1 + x} d x c a l c u l a t e

U_{n} + U_{n + 1}

Commented by jagoll last updated on 15/Jan/20

U_{n} + U_{n + 1} = \int \underset{0}{\overset{1}{}} \frac{x^{n} + x^{n + 1}}{1 + x} d x

= \underset{0}{\int^{1}} \frac{x^{n} (1 + x)}{1 + x} d x = \frac{x^{n + 1}}{n + 1} ∣_{0}^{1}

= \frac{1}{n + 1}

Commented by mr W last updated on 16/Jan/20

U_{n} = ?

Commented by john santu last updated on 16/Jan/20

U_{n} = \int_{0}^{1} \frac{x^{n}}{1 + x} d x = \int_{1}^{2} \frac{{(u - 1)}^{n}}{u} d u

= \int_{1}^{2} \frac{\underset{r = 0}{\sum^{n}} C_{r}^{n} u^{n - r} {(- 1)}^{r}}{u} d u

= \int_{1}^{2} \underset{r = 0}{\sum^{n}} C_{r}^{n} u^{n - r - 1} {(- 1)}^{r} d u

= \underset{r = 0}{\sum^{n}} C_{r}^{n} {(- 1)}^{r} \times \frac{u^{n - r}}{n - r} ∣_{1}^{2}