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

Evaluer  ∫((1−logx)/(1+x))dx

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

Answered by ARUNG_Brandon_MBU last updated on 29/Oct/22

∫((1−logx)/(1+x))dx  =log(1+x)−Σ_(n=0) ^∞ (−1)^n ∫x^n logxdx  =log(1+x)−Σ_(n≥0) (−1)^n (((x^(n+1) logx)/(n+1))−(1/(n+1))∫x^n dx)  =log(1+x)−Σ_(n≥0) (−1)^n (((x^(n+1) logx)/(n+1))−(x^(n+1) /((n+1)^2 )))

$$\int\frac{\mathrm{1}−\mathrm{log}{x}}{\mathrm{1}+{x}}{dx} \\ $$$$=\mathrm{log}\left(\mathrm{1}+{x}\right)−\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \int{x}^{{n}} \mathrm{log}{xdx} \\ $$$$=\mathrm{log}\left(\mathrm{1}+{x}\right)−\underset{{n}\geqslant\mathrm{0}} {\sum}\left(−\mathrm{1}\right)^{{n}} \left(\frac{{x}^{{n}+\mathrm{1}} \mathrm{log}{x}}{{n}+\mathrm{1}}−\frac{\mathrm{1}}{{n}+\mathrm{1}}\int{x}^{{n}} {dx}\right) \\ $$$$=\mathrm{log}\left(\mathrm{1}+{x}\right)−\underset{{n}\geqslant\mathrm{0}} {\sum}\left(−\mathrm{1}\right)^{{n}} \left(\frac{{x}^{{n}+\mathrm{1}} \mathrm{log}{x}}{{n}+\mathrm{1}}−\frac{{x}^{{n}+\mathrm{1}} }{\left({n}+\mathrm{1}\right)^{\mathrm{2}} }\right) \\ $$

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

good thank you

$${good}\:{thank}\:{you} \\ $$

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