Question Number 131884 by Study last updated on 09/Feb/21 $${log}_{\mathrm{2}} {x}+{log}_{\mathrm{3}} {x}=\mathrm{1}\:\:\:\:\:\:\:{x}=? \\ $$ Answered by EDWIN88 last updated on 09/Feb/21 $$\:\frac{\mathrm{ln}\:\mathrm{x}}{\mathrm{ln}\:\mathrm{2}}\:+\:\frac{\mathrm{ln}\:\mathrm{x}}{\mathrm{ln}\:\mathrm{3}}\:=\:\mathrm{1}\: \\ $$$$\:\mathrm{ln}\:\mathrm{x}\:\left(\frac{\mathrm{1}}{\mathrm{ln}\:\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{ln}\:\mathrm{3}}\right)=\mathrm{1} \\…
Question Number 66351 by mathmax by abdo last updated on 12/Aug/19 $${let}\:{I}_{{n}} =\int_{\mathrm{0}} ^{\infty} \:\:\frac{{e}^{{nt}} }{\left(\mathrm{1}+{e}^{{t}} \right)^{{n}+\mathrm{1}} }{dt}\:\:\:\:\:\left({n}\:{from}\:{N}^{\bigstar} \right) \\ $$$$\left.\right){prove}\:{the}\:{existence}\:{of}\:{I}_{{n}} \\ $$$$\left.\mathrm{2}\right){find}\:{lim}_{{n}\rightarrow+\infty} \:\:\:{I}_{{n}} \\…
Question Number 131887 by Algoritm last updated on 09/Feb/21 Answered by SEKRET last updated on 09/Feb/21 $$\:\boldsymbol{\mathrm{Leybnist}}\:\:\:\boldsymbol{\mathrm{formula}} \\ $$$$\:\:\boldsymbol{\mathrm{u}}=\:\boldsymbol{\mathrm{e}}^{−\mathrm{2}\boldsymbol{\mathrm{x}}} \:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{u}}^{\boldsymbol{\mathrm{n}}} =\left(−\mathrm{2}\right)^{\boldsymbol{\mathrm{n}}} \centerdot\boldsymbol{\mathrm{e}}^{−\mathrm{2}\boldsymbol{\mathrm{x}}} \\ $$$$\:\boldsymbol{\mathrm{v}}=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−\boldsymbol{\mathrm{x}}}}\:\:\:\:\:\:\boldsymbol{\mathrm{v}}^{\boldsymbol{\mathrm{n}}} =\frac{\left(\mathrm{2}\boldsymbol{\mathrm{n}}−\mathrm{1}\right)!!}{\mathrm{2}^{\boldsymbol{\mathrm{n}}}…
Question Number 66348 by mathmax by abdo last updated on 12/Aug/19 $${find}\:{nature}\:{of}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{dx}}{{e}^{{x}} −{cosx}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 66349 by mathmax by abdo last updated on 12/Aug/19 $${study}\:{the}\:{convergence}\:{of}\:\int_{\mathrm{1}} ^{+\infty} \:\:\frac{{arctan}\left({x}−\mathrm{1}\right)}{\left({x}^{\mathrm{2}} −\mathrm{1}\right)^{\frac{\mathrm{4}}{\mathrm{3}}} }{dx} \\ $$ Commented by mathmax by abdo last updated…
Question Number 66346 by mathmax by abdo last updated on 12/Aug/19 $${find}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{t}^{\mathrm{7}} }{{t}^{\mathrm{16}} \:+\mathrm{1}}{dt} \\ $$ Commented by mathmax by abdo last updated…
Question Number 131880 by Algoritm last updated on 09/Feb/21 Answered by SEKRET last updated on 09/Feb/21 $$\:\boldsymbol{\mathrm{x}}=\frac{\mathrm{5}}{\mathrm{2}} \\ $$ Commented by Algoritm last updated on…
Question Number 66347 by mathmax by abdo last updated on 12/Aug/19 $${let}\:{I}_{{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{x}^{\mathrm{2}{n}+\mathrm{1}} {ln}\left({x}\right)}{{x}^{\mathrm{2}} −\mathrm{1}}{dx} \\ $$$$\left.\mathrm{1}\right)\:{prove}\:{the}\:{existence}\:{of}\:{I}_{{n}} \\ $$$$\left.\mathrm{2}\right){calculate}\:{I}_{{n}+\mathrm{1}} −{I}_{{n}} \\ $$$$\left.\mathrm{3}\left.\right){prove}\:{thst}\:{x}\in\right]\mathrm{0},\mathrm{1}\left[\:\Rightarrow\mathrm{0}<\frac{{xlnx}}{{x}^{\mathrm{2}} −\mathrm{1}}<\frac{\mathrm{1}}{\mathrm{2}}\right.…
Question Number 66344 by mathmax by abdo last updated on 12/Aug/19 $${let}\:{f}_{{n}} \left({x}\right)=\frac{\mathrm{1}}{\left(\mathrm{1}+{x}^{{n}} \right)^{\mathrm{1}+\frac{\mathrm{1}}{{n}}} }\:\:\:{defined}\:{on}\:\left[\mathrm{0},\mathrm{1}\right] \\ $$$$\left.\mathrm{1}\right){prove}\:{that}\:{f}_{{n}} \rightarrow^{{cs}} \:\:{to}\:{a}\:{function}\:{f}\:{on}\left[\mathrm{0},\mathrm{1}\right] \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:{I}_{{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} {f}_{{n}} \left({x}\right){dx}…
Question Number 131882 by Eric002 last updated on 09/Feb/21 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}}{{x}+\sqrt[{\mathrm{4}}]{{x}+\mathrm{1}}−\mathrm{1}} \\ $$ Answered by liberty last updated on 09/Feb/21 $$\:\mathrm{L}'\mathrm{H}\ddot {\mathrm{o}pital}\:\mathrm{L}=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\left[\:\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}\:\sqrt[{\mathrm{4}}]{\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{3}} }}}\:\right]=\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}}}=\:\frac{\mathrm{4}}{\mathrm{5}} \\…