Question Number 133957 by mathmax by abdo last updated on 25/Feb/21 $$\left.\mathrm{1}\right)\mathrm{decompose}\:\:\mathrm{F}\left(\mathrm{x}\right)=\frac{\mathrm{1}}{\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{5}} \left(\mathrm{2x}−\mathrm{1}\right)^{\mathrm{4}} } \\ $$$$\left.\mathrm{2}\right)\:\mathrm{find}\:\int_{\mathrm{1}} ^{\infty} \:\mathrm{F}\left(\mathrm{x}\right)\mathrm{dx} \\ $$ Answered by Olaf last updated…
Question Number 133951 by liberty last updated on 25/Feb/21 $$\:\mathcal{V}\:=\:\int\:\mathrm{ln}\:\left(\mathrm{x}+\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\:\right)\:\mathrm{dx}\: \\ $$ Answered by mathmax by abdo last updated on 25/Feb/21 $$\Phi\:=\int\:\mathrm{ln}\left(\mathrm{x}+\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\right)\mathrm{dx}\:\mathrm{we}\:\mathrm{do}\:\mathrm{the}\:\mathrm{changement}\:\mathrm{x}=\mathrm{sht}\:\Rightarrow \\…
Question Number 68409 by mathmax by abdo last updated on 10/Sep/19 $${calculate}\:\int_{\mathrm{0}} ^{+\infty} \:\:\frac{{arctan}\left({x}^{\mathrm{2}} \right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$ Commented by mathmax by abdo last updated…
Question Number 133943 by mnjuly1970 last updated on 25/Feb/21 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…….{advanced}\:\:\:\:\:{calculus}…… \\ $$$$\:\:\:\:\:\:{prove}\:\:{that}: \\ $$$$\:\:\:\:\:\:\boldsymbol{\phi}=\:\:\int_{\mathrm{0}} ^{\:\infty} \frac{{cos}\left({x}^{\mathrm{2}} \right)−{cos}\left({x}\right)}{{x}}{dx}=\frac{\gamma}{\mathrm{2}} \\ $$$$\:\:\:\gamma:\:{euler}−{mascheroni}\:{constant}… \\ $$ Terms of Service Privacy…
Question Number 133938 by Algoritm last updated on 25/Feb/21 Answered by mathmax by abdo last updated on 25/Feb/21 $$\mathrm{I}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{ln}\left(−\mathrm{lnx}\right)}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx}\:\mathrm{changement}\:−\mathrm{lnx}=\mathrm{t}\:\mathrm{give}\:\mathrm{x}=\mathrm{e}^{−\mathrm{t}} \:\Rightarrow \\ $$$$\mathrm{I}=\int_{\mathrm{0}}…
Question Number 133911 by liberty last updated on 25/Feb/21 $$\mathcal{A}=\int\:\frac{\mathrm{cos}\:\mathrm{x}}{\mathrm{sin}\:\mathrm{5x}+\mathrm{sin}\:\mathrm{x}}\:\mathrm{dx}\:? \\ $$ Answered by Ar Brandon last updated on 25/Feb/21 $$\mathcal{A}=\int\frac{\mathrm{cosx}}{\mathrm{sin5x}+\mathrm{sinx}}\mathrm{dx}=\int\frac{\mathrm{cosx}}{\mathrm{2sin3xcos2x}}\mathrm{dx} \\ $$$$\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{cosx}\:\mathrm{dx}}{\left(\mathrm{3sinx}−\mathrm{4sin}^{\mathrm{3}} \mathrm{x}\right)\left(\mathrm{1}−\mathrm{2sin}^{\mathrm{2}} \mathrm{x}\right)}…
Question Number 133910 by Ar Brandon last updated on 25/Feb/21 $$\:\:\:\:\mathcal{D}\acute {\mathrm{e}montrer}\:\mathrm{que}; \\ $$$$\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{1}+\mathrm{n}^{\mathrm{4}} }=\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\pi}{\:\sqrt{\mathrm{2}}}\:\frac{\mathrm{sin}\left(\frac{\pi}{\:\sqrt{\mathrm{2}}}\right)\mathrm{cosh}\left(\frac{\pi}{\:\sqrt{\mathrm{2}}}\right)+\mathrm{sinh}\left(\frac{\pi}{\:\sqrt{\mathrm{2}}}\right)\mathrm{cos}\left(\frac{\pi}{\:\sqrt{\mathrm{2}}}\right)}{\mathrm{sinh}^{\mathrm{2}} \left(\frac{\pi}{\:\sqrt{\mathrm{2}}}\right)+\mathrm{sin}^{\mathrm{2}} \left(\frac{\pi}{\:\sqrt{\mathrm{2}}}\right)}\right] \\ $$ Terms of Service…
Question Number 133907 by mnjuly1970 last updated on 25/Feb/21 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 68370 by mhmd last updated on 09/Sep/19 Answered by MJS last updated on 09/Sep/19 $$\int\frac{{dt}}{\:\sqrt{\mathrm{e}^{{t}} −\mathrm{1}}}= \\ $$$$\:\:\:\:\:\left[{u}=\sqrt{\mathrm{e}^{{t}} −\mathrm{1}}\:\rightarrow\:{dt}=\frac{\mathrm{2}\sqrt{\mathrm{e}^{{t}} −\mathrm{1}}}{\mathrm{e}^{{t}} }{du}\right] \\ $$$$=\mathrm{2}\int\frac{{du}}{{u}^{\mathrm{2}}…
Question Number 2818 by prakash jain last updated on 27/Nov/15 $$\mathrm{show}\:\mathrm{that} \\ $$$$\Gamma'\left(\mathrm{1}\right)=−\gamma \\ $$$$\Gamma\:\mathrm{gamma}\:\mathrm{function} \\ $$$$\gamma=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left[\mathrm{H}_{{n}} −\mathrm{ln}\:{n}\right] \\ $$ Answered by 123456 last…