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Question Number 179494 by a.lgnaoui last updated on 29/Oct/22

$$Evaluer$$$$\int \frac{1-\mathrm{logx}}{1+\mathrm{x}}\mathrm{dx}$$

Answered by ARUNG_Brandon_MBU last updated on 29/Oct/22

$$\int \frac{1-\log x}{1+x}dx$$$$=\log (1+x)-\underset{n=0}{\sum ^{\mathrm{\infty }}}{(-1)}^n\int x^n\log xdx$$$$=\log (1+x)-\sum _{n⩾0}{(-1)}^n(\frac{x^{n+1}\log x}{n+1}-\frac{1}{n+1}\int x^{n}dx)$$$$=\log (1+x)-\sum _{n⩾0}{(-1)}^n(\frac{x^{n+1}\log x}{n+1}-\frac{x^{n+1}}{{(n+1)}^2})$$

Commented by a.lgnaoui last updated on 30/Oct/22

$$goodthankyou$$

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