Question Number 134289 by bramlexs22 last updated on 02/Mar/21 $$\mathrm{I}=\int\:\frac{\mathrm{x}^{\mathrm{n}} }{\:\sqrt{\mathrm{ax}+\mathrm{b}}}\:\mathrm{dx} \\ $$$$\mathrm{H}=\int\:\frac{\mathrm{x}^{\mathrm{4}} }{\:\sqrt{\mathrm{2x}+\mathrm{1}}}\:\mathrm{dx} \\ $$ Answered by EDWIN88 last updated on 02/Mar/21 $$\mathrm{I}_{\mathrm{n}} =\:\frac{\mathrm{2x}^{\mathrm{n}}…
Question Number 134291 by Lordose last updated on 02/Mar/21 $$ \\ $$$$\:\boldsymbol{\mathrm{Prove}}\:\:\:\int_{\mathrm{0}} ^{\:\infty} \frac{\mathrm{x}^{\mathrm{a}} }{\mathrm{1}+\mathrm{e}^{\mathrm{x}} }\mathrm{dx}\:=\:\left(\mathrm{1}−\mathrm{2}^{−\mathrm{a}} \right)\boldsymbol{\zeta}\left(\mathrm{a}+\mathrm{1}\right)\boldsymbol{\Gamma}\left(\mathrm{a}+\mathrm{1}\right) \\ $$$$ \\ $$ Answered by Dwaipayan Shikari…
Question Number 134272 by mnjuly1970 last updated on 01/Mar/21 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{nice}\:\:\:\mathrm{calculus} \\ $$$$\:\:\:\:\mathrm{prove}\:\mathrm{that}:\::\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:{i}\:::\:\: =\int_{\mathrm{0}} ^{\:\infty} \frac{{cos}\left(\pi{x}\right)}{{e}^{\mathrm{2}\pi\sqrt{{x}}\:} −\mathrm{1}}\:{dx}=\frac{\mathrm{2}−\sqrt{\mathrm{2}}\:}{\mathrm{8}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{ii}::\:{compute}:\:\:\underset{{n}=−\infty} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{4}} +\mathrm{9}{n}^{\mathrm{2}} +\mathrm{10}}\:=? \\…
Question Number 3190 by Filup last updated on 07/Dec/15 $$\mathrm{How}\:\mathrm{do}\:\mathrm{you}\:\mathrm{find}\:\mathrm{the}\:\mathrm{length}\:\mathrm{of} \\ $$$$\mathrm{a}\:\mathrm{curve}\:\mathrm{between}\:{f}\left({a}\right)\:\mathrm{and}\:{f}\left({b}\right)? \\ $$ Commented by 123456 last updated on 07/Dec/15 $$\mathrm{wath}\:\mathrm{you}\:\mathrm{mean}\:\mathrm{by}\:\mathrm{lenght}\:\mathrm{of}\:\mathrm{curve}? \\ $$$$\mathrm{if}\:\mathrm{i}\:\mathrm{undestand}\:\mathrm{right},\:\mathrm{if}\:{f}\:\mathrm{is}\:\mathrm{continuous} \\…
Question Number 68699 by Rio Michael last updated on 15/Sep/19 $$\int\:{ln}\left({x}\:+\:\mathrm{4}\right)\:{dx}\:= \\ $$ Commented by mathmax by abdo last updated on 15/Sep/19 $${let}\:{I}\:=\int{ln}\left({x}+\mathrm{4}\right){dx}\:\:{changement}\:{x}+\mathrm{4}={t}\:{give} \\ $$$${I}\:=\int{lnt}\:{dt}\:={tln}\left({t}\right)−{t}\:+{c}\:=\left({x}+\mathrm{4}\right){ln}\left({x}+\mathrm{4}\right)−{x}−\mathrm{4}\:+{c}…
Question Number 134239 by ruwedkabeh last updated on 01/Mar/21 $${can}\:{i}\:{ask}\:{for}\:{some}\:{help}? \\ $$$${how}\:{to}\:{prove}\:{this}? \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}<\underset{\mathrm{0}} {\overset{\frac{\mathrm{1}}{\mathrm{2}}} {\int}}\frac{{dx}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{3}} }}<\frac{\pi}{\mathrm{6}} \\ $$ Commented by Dwaipayan Shikari last updated…
Question Number 134227 by Raxreedoroid last updated on 01/Mar/21 $$\mathrm{I}\:\:\mathrm{struck}\:\mathrm{upon}\:\mathrm{this} \\ $$$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\mathrm{16}^{{n}} =\frac{−\mathrm{1}}{\mathrm{15}} \\ $$$$\int_{\mathrm{0}} ^{\:\pi} \mathrm{cos}^{\mathrm{4}} \:{x}\:\mathrm{sin}\:{x}\:{dx}=\frac{−\mathrm{1}}{\mathrm{5}}\left[\mathrm{cos}^{\mathrm{5}} \:{x}\right]_{\mathrm{0}} ^{\pi} =\mathrm{0}.\mathrm{4} \\ $$$$\mathrm{in}\:\mathrm{another}\:\mathrm{way}…
Question Number 134219 by Raxreedoroid last updated on 01/Mar/21 $$\int_{\mathrm{0}} ^{\:\pi} \mathrm{cos}^{\mathrm{4}} \:{x}\:\mathrm{sin}\:{x}\:{dx}=? \\ $$ Answered by liberty last updated on 01/Mar/21 $$=−\left[\frac{\mathrm{1}}{\mathrm{5}}\mathrm{cos}\:^{\mathrm{5}} \mathrm{x}\:\right]_{\mathrm{0}} ^{\pi}…
Question Number 68660 by aliesam last updated on 14/Sep/19 $$\int{e}^{{x}+{e}^{{x}} } \:{dx} \\ $$ Commented by aliesam last updated on 14/Sep/19 $${perfect} \\ $$ Commented…
Question Number 134185 by mnjuly1970 last updated on 28/Feb/21 $$\:\:\:\:\:\:\:\:\:\:\:\:…..{advanced}\:\:\:{calculus}…. \\ $$$$\:\:\:\:{prove}\:\:{that}:: \\ $$$$\:\:\:\:\:{i}:\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \frac{{cos}\left({tan}\left({x}\right)−{x}\right)}{{cos}\left({x}\right)}{dx}=\frac{\pi}{{e}} \\ $$$$\:\:\:\:{ii}:\underset{{n}=\mathrm{2}} {\overset{\infty} {\prod}}{e}\left(\mathrm{1}−\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\right)^{{n}^{\mathrm{2}} } =\frac{\pi}{{e}\sqrt{{e}}} \\ $$$$\:\:…