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Author: Tinku Tara

Question-130188

Question Number 130188 by SEKRET last updated on 23/Jan/21 Answered by Lordose last updated on 23/Jan/21 $$ \\ $$$$\int_{\mathrm{0}} ^{\:\infty} \frac{\mathrm{x}^{\frac{\mathrm{5}}{\mathrm{4}}−\mathrm{1}} }{\left(\mathrm{1}+\mathrm{x}\right)^{\frac{\mathrm{5}}{\mathrm{4}}+\frac{\mathrm{3}}{\mathrm{4}}} }\mathrm{dx}\:=\:\boldsymbol{\beta}\left(\frac{\mathrm{5}}{\mathrm{4}},\frac{\mathrm{3}}{\mathrm{4}}\right)\:=\:\frac{\boldsymbol{\Gamma}\left(\frac{\mathrm{5}}{\mathrm{4}}\right)\boldsymbol{\Gamma}\left(\frac{\mathrm{3}}{\mathrm{4}}\right)}{\boldsymbol{\Gamma}\left(\mathrm{2}\right)}\:=\:\frac{\boldsymbol{\pi}}{\mathrm{2}\sqrt{\mathrm{2}}}\:\:\:\:\:\:\:\:\:\:\:\: \\ $$…

if-Z-is-any-boint-on-the-circle-Z-1-1-prove-that-arg-Z-1-2arg-Z-2-3-arg-Z-2-Z-

Question Number 130186 by mohammad17 last updated on 23/Jan/21 $${if}\:{Z}\:{is}\:{any}\:{boint}\:{on}\:{the}\:{circle}\:\mid{Z}−\mathrm{1}\mid=\mathrm{1} \\ $$$$ \\ $$$${prove}\:{that}\:{arg}\left({Z}−\mathrm{1}\right)=\mathrm{2}{arg}\left({Z}\right)=\frac{\mathrm{2}}{\mathrm{3}}{arg}\left({Z}^{\mathrm{2}} −{Z}\right) \\ $$ Commented by mohammad17 last updated on 23/Jan/21 $$?????…

Question-130181

Question Number 130181 by Adel last updated on 23/Jan/21 Answered by Olaf last updated on 23/Jan/21 $$ \\ $$$$\underset{{x}\rightarrow\mathrm{3}} {\mathrm{lim}}\frac{\Gamma\left({x}+\mathrm{1}\right)−\mathrm{6}}{{xx}−\mathrm{33}} \\ $$$$=\:\underset{{x}\rightarrow\mathrm{3}} {\mathrm{lim}}\frac{\Gamma\left({x}+\mathrm{1}\right)−\mathrm{6}}{\mathrm{11}\left({x}−\mathrm{3}\right)} \\ $$$$=\:\underset{{x}\rightarrow\mathrm{3}}…

dx-e-x-x-

Question Number 64642 by mmkkmm000m last updated on 19/Jul/19 $$\int\left({dx}\right)/{e}^{{x}} +{x} \\ $$ Commented by mathmax by abdo last updated on 20/Jul/19 $${let}\:{I}\:=\int\:\:\:\frac{{dx}}{{x}+{e}^{{x}} }\:\Rightarrow{I}\:=\int\:\:\:\frac{{e}^{−{x}} }{{xe}^{−{x}}…

lim-x-6x-8x-2-2x-3-4x-5-x-6-x-3-4x-5-1-6-x-

Question Number 130176 by liberty last updated on 23/Jan/21 $$\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{6x}+\sqrt{\mathrm{8x}^{\mathrm{2}} −\sqrt{\mathrm{2x}^{\mathrm{3}} +\sqrt{\mathrm{4x}^{\mathrm{5}} }}}}{\:\sqrt[{\mathrm{6}}]{\mathrm{x}^{\mathrm{6}} −\mathrm{x}^{\mathrm{3}} +\sqrt{\mathrm{4x}^{\mathrm{5}} }}\:+\:\mathrm{x}}=? \\ $$ Answered by EDWIN88 last updated on…