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

$$let{U}_n={\int }_0^1\frac{x^n}{1+x}dxcalculate$$$$U_n+U_{n+1}$$

Commented by jagoll last updated on 15/Jan/20

$$U_n+U_{n+1}=\int \underset{0}{\stackrel{1}{}}\frac{x^n+x^{n+1}}{1+x}dx$$$$=\underset{0}{\stackrel{1}{\int }}\frac{x^n(1+x)}{1+x}dx=\frac{x^{n+1}}{n+1}{\mid }_0^1$$$$=\frac{1}{n+1}$$

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

$$U_n=?$$

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

$$U_n={\int }_0^1\frac{x^n}{1+x}dx={\int }_1^2\frac{{(u-1)}^n}{u}du$$$$={\int }_1^2\frac{\underset{r=0}{\sum ^n}C{}_r^n{u}^{n-r}{(-1)}^r}{u}du$$$$={\int }_1^2\underset{r=0}{\sum ^n}{C}_r^n{u}^{n-r-1}{(-1)}^{r}du$$$$=\underset{r=0}{\sum ^n}{C}_r^n{(-1)}^r\times \frac{u^{n-r}}{n-r}{\mid }_1^2$$

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