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Question Number 113714 by mathdave last updated on 14/Sep/20

Answered by Olaf last updated on 15/Sep/20

$${\mathrm{I}}_n={\int }_0^1\frac{x^n\ln (1+x)}{1+x}dx$$$${\mathrm{I}}_0={\int }_0^1\frac{\ln (1+x)dx}{1+x}$$$${\mathrm{I}}_0={\left[\frac{1}{2}{\ln }^2(1+x)\right]}_0^1=\frac{1}{2}{\ln }^{2}2$$$${\mathrm{I}}_{n+1}+{\mathrm{I}}_n={\int }_0^1\frac{x^{n+1}+x^n}{1+x}\ln (1+x)dx$$$${\mathrm{I}}_{n+1}+{\mathrm{I}}_n={\int }_0^1{x}^n\ln (1+x)dx$$$$\frac{1}{1+x}=\underset{k=0}{\sum ^{\mathrm{\infty }}}{(-1)}^k{x}^k$$$$\ln (1+x)=\underset{k=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^k{x}^{k+1}}{k+1},\mid x\mid <1$$$${\mathrm{I}}_{n+1}+{\mathrm{I}}_n={\int }_0^1{x}^n\underset{k=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^k{x}^{k+1}}{k+1}dx$$$${\mathrm{I}}_{n+1}+{\mathrm{I}}_n={\int }_0^1\underset{k=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^k{x}^{n+k+1}}{k+1}dx$$$${\mathrm{I}}_{n+1}+{\mathrm{I}}_n={\left[\underset{k=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^k{x}^{n+k+1}}{(k+1)(n+k+1)}\right]}_0^1$$$${\mathrm{I}}_{n+1}+{\mathrm{I}}_n=\underset{k=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^k}{(k+1)(n+k+1)}$$$$\frac{1}{(k+1)(n+k+1)}=\frac{1}{n}(\frac{1}{k+1}-\frac{1}{n+k+1})$$$${\mathrm{I}}_{n+1}+{\mathrm{I}}_n=\frac{1}{n}\underset{k=0}{\sum ^{\mathrm{\infty }}}{(-1)}^k(\frac{1}{k+1}-\frac{1}{n+k+1})$$$$...$$

Commented by mathdave last updated on 15/Sep/20

$$welldonebutwaitoooifintegrate$$$${\int }_0^1{x}^{k+n+1}dxthissupposedtogave$$$$\frac{x^{k+n+2}}{k+n+2}$$

Commented by Tawa11 last updated on 06/Sep/21

$$\mathrm{great}\mathrm{sir}$$

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