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Author: Tinku Tara

1-x-dx-

Question Number 64213 by Tony Lin last updated on 15/Jul/19 $$\int\frac{\mathrm{1}}{{x}!}{dx}=? \\ $$ Commented by MJS last updated on 15/Jul/19 $$\frac{\mathrm{1}}{{x}!}\:\mathrm{is}\:\mathrm{defined}\:\mathrm{for}\:{x}\in\mathbb{N}\:\Rightarrow\:\mathrm{it}'\mathrm{s}\:\mathrm{not}\:\mathrm{continuous} \\ $$$$\mathrm{for}\:{x}\in\mathbb{R}\:\Rightarrow\:\mathrm{it}'\mathrm{s}\:\mathrm{not}\:\mathrm{integrable} \\ $$…

N-3-2cos-x-2-3cos-x-dx-

Question Number 129746 by bramlexs22 last updated on 18/Jan/21 $$\:\mathrm{N}\:=\:\int\:\frac{\mathrm{3}+\mathrm{2cos}\:\mathrm{x}}{\mathrm{2}+\mathrm{3cos}\:\mathrm{x}}\:\mathrm{dx}\: \\ $$ Answered by liberty last updated on 18/Jan/21 $$\:\mathrm{N}=\:\int\:\frac{\frac{\mathrm{2}}{\mathrm{3}}\left(\mathrm{2}+\mathrm{3cos}\:\mathrm{x}\right)+\frac{\mathrm{8}}{\mathrm{3}}}{\mathrm{2}+\mathrm{3cos}\:\mathrm{x}}\:\mathrm{dx}\: \\ $$$$\:\mathrm{N}=\:\frac{\mathrm{2}}{\mathrm{3}}\mathrm{x}\:+\:\frac{\mathrm{8}}{\mathrm{3}}\int\:\frac{\mathrm{dx}}{\mathrm{2}+\mathrm{3}\left(\mathrm{2cos}\:^{\mathrm{2}} \frac{\mathrm{x}}{\mathrm{2}}−\mathrm{1}\right)} \\ $$$$\:\mathrm{N}=\:\frac{\mathrm{2}}{\mathrm{3}}\mathrm{x}+\frac{\mathrm{8}}{\mathrm{3}}\int\:\frac{\mathrm{dx}}{\mathrm{6cos}\:^{\mathrm{2}}…

M-is-the-midpoint-of-segment-AB-Prove-that-for-every-point-P-in-space-PM-PA-PB-2-

Question Number 64203 by Prithwish sen last updated on 15/Jul/19 $$\mathrm{M}\:\mathrm{is}\:\mathrm{the}\:\mathrm{midpoint}\:\mathrm{of}\:\mathrm{segment}\:\mathrm{AB}.\:\mathrm{Prove} \\ $$$$\mathrm{that},\mathrm{for}\:\mathrm{every}\:\mathrm{point}\:\mathrm{P}\:\mathrm{in}\:\mathrm{space}, \\ $$$$\mid\mathrm{PM}\mid\:\leqslant\:\frac{\mid\mathrm{PA}\mid\:+\mid\mathrm{PB}\mid}{\mathrm{2}} \\ $$ Commented by Prithwish sen last updated on 16/Jul/19…

Find-all-positive-solutions-x-y-z-1-x-3-y-3-z-3-xyz-x-4-y-4-z-4-1-

Question Number 64201 by Prithwish sen last updated on 15/Jul/19 $$\mathrm{Find}\:\mathrm{all}\:\mathrm{positive}\:\mathrm{solutions}\: \\ $$$$\mathrm{x}+\mathrm{y}+\mathrm{z}\:=\:\mathrm{1} \\ $$$$\mathrm{x}^{\mathrm{3}} +\mathrm{y}^{\mathrm{3}} +\:\mathrm{z}^{\mathrm{3}} \:+\:\mathrm{xyz}\:=\:\mathrm{x}^{\mathrm{4}} \:+\:\mathrm{y}^{\mathrm{4}} \:+\:\mathrm{z}^{\mathrm{4}} \:+\mathrm{1} \\ $$ Answered by…

Question-64200

Question Number 64200 by aliesam last updated on 15/Jul/19 Commented by mathmax by abdo last updated on 15/Jul/19 $${let}\:{I}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{x}−\mathrm{1}}{{lnx}}{dx}\:{changement}\:{lnx}=−{t}\:{give}\:{x}={e}^{−{t}} \\ $$$${I}\:=−\int_{\mathrm{0}} ^{\infty} \:\:\frac{{e}^{−{t}}…