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Author: Tinku Tara

Question-205237

Question Number 205237 by universe last updated on 13/Mar/24 Answered by Berbere last updated on 13/Mar/24 $${n}^{\mathrm{2}} +{x}^{\mathrm{2}} \geqslant{n}^{\mathrm{2}} \\ $$$$\frac{{x}}{\mathrm{1}+{x}}\leqslant\mathrm{1}\Rightarrow\frac{{nx}\mathrm{tan}^{−\mathrm{1}} \left({x}\right)}{\left(\mathrm{1}+{x}\right)\left({n}^{\mathrm{2}} +{x}^{\mathrm{2}} \right)}\leqslant{n}.\mathrm{1}.\frac{\mathrm{tan}^{−\mathrm{1}} \left({x}\right)}{{n}^{\mathrm{2}}…

pls-how-to-calculate-this-1-2-1-ln-x-1-x-dx-

Question Number 205264 by pticantor last updated on 13/Mar/24 $$\boldsymbol{{pls}}\:\boldsymbol{{how}}\:\boldsymbol{{to}}\:\boldsymbol{{calculate}}\:\boldsymbol{{this}}? \\ $$$$\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} \frac{\boldsymbol{{ln}}\left(\boldsymbol{{x}}+\mathrm{1}\right)}{\boldsymbol{{x}}}\boldsymbol{{dx}} \\ $$ Answered by Berbere last updated on 13/Mar/24 $$=−\left(−\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}}…

nature-of-the-serie-n-1-ln-n-n-

Question Number 205262 by mathzup last updated on 13/Mar/24 $${nature}\:{of}\:{the}\:{serie}\:\sum_{{n}\geqslant\mathrm{1}} \:\frac{{ln}\left({n}\right)}{{n}} \\ $$ Answered by Berbere last updated on 13/Mar/24 $$\forall{n}\geqslant\mathrm{2}\:{ln}\left({n}\right)\geqslant{ln}\left(\mathrm{2}\right)>\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{{ln}\left({n}\right)}{{n}}\geqslant\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{{n}}\rightarrow+\infty\:{serie}\:{dv}…

0-pi-x-2-cos-2-x-xsin-x-cos-x-1-1-xsin-x-2-dx-

Question Number 205248 by universe last updated on 13/Mar/24 $$\:\:\:\int_{\mathrm{0}} ^{\pi} \:\frac{{x}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \left({x}\right)−{x}\mathrm{sin}\left({x}\right)−\mathrm{cos}\left({x}\right)−\mathrm{1}}{\left(\mathrm{1}+{x}\mathrm{sin}\left({x}\right)\right)^{\mathrm{2}} }{dx} \\ $$ Answered by Berbere last updated on 13/Mar/24 $$\left(\frac{{f}\left({x}\right)}{\mathrm{1}+{xsin}\left({x}\right)}+{g}\left({x}\right)\right)^{'}…