Question Number 198497 by universe last updated on 21/Oct/23 $$\:\:\:\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{e}^{\mathrm{at}} −\mathrm{e}^{−\mathrm{at}} }{\mathrm{e}^{\pi\mathrm{t}} −\mathrm{e}^{−\pi\mathrm{t}} }\:\mathrm{dt} \\ $$ Commented by Hridiana last updated on 21/Oct/23…
Question Number 198496 by mnjuly1970 last updated on 21/Oct/23 $$ \\ $$$$\:\:\:\:\:\:\:\:\:{A}\:{nice}\:\:{series} \\ $$$$\:\:{If}\:,\:\:\Omega\:=\:\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\:\frac{\:\mathrm{1}}{{n}^{\:\mathrm{2}} \:+{n}\:−\mathrm{1}}\:=\frac{\:\pi\:{tan}\left(\:{a}\pi\:\right)}{\:{b}} \\ $$$$\:\:\:\:\:\Rightarrow\:{find}\:{the}\:{value}\:{of}\:\:\:\frac{{b}}{{a}}\:=\:? \\ $$$$\:\:\:\:\:\:\:\:\: \\ $$ Commented by…
Question Number 198535 by Hridiana last updated on 21/Oct/23 $${t}=\mathrm{4}\left\{{t}=\mathrm{4}\left\{{t}=\mathrm{4}…\right\}\right\} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 198420 by Mingma last updated on 19/Oct/23 Answered by Frix last updated on 19/Oct/23 $$\int\frac{{x}^{\mathrm{2}} }{\left(\mathrm{3}+\mathrm{4}{x}−\mathrm{4}{x}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} }{dx}\:\underset{{dx}=−\frac{\left(\mathrm{2}{x}−\mathrm{1}\right)\sqrt{\mathrm{3}+\mathrm{4}{x}−\mathrm{4}{x}^{\mathrm{2}} }}{\mathrm{4}{t}}} {\overset{{t}=\frac{\mathrm{2}+\sqrt{\mathrm{3}+\mathrm{4}{x}−\mathrm{4}{x}^{\mathrm{2}} }}{\mathrm{2}{x}−\mathrm{1}}} {=}}\: \\…
Please-Help-S-x-2-dydz-y-2-dzdx-2z-xy-x-y-dxdy-where-S-is-the-surface-of-the-cube-0-x-1-0-y-1-0-z-1-
Question Number 198419 by Nimnim111118 last updated on 19/Oct/23 $$\:\:{Please}\:{Help}… \\ $$$$\:\:\int\underset{{S}} {\int}{x}^{\mathrm{2}} {dydz}+{y}^{\mathrm{2}} {dzdx}+\mathrm{2}{z}\left({xy}−{x}−{y}\right){dxdy}\:{where} \\ $$$$\:\:\:{S}\:{is}\:{the}\:{surface}\:{of}\:{the}\:{cube}.\:\mathrm{0}\leqslant{x}\leqslant\mathrm{1},\:\mathrm{0}\leqslant{y}\leqslant\mathrm{1}, \\ $$$$\:\:\:\:\mathrm{0}\leqslant{z}\leqslant\mathrm{1} \\ $$$$ \\ $$ Terms of…
Question Number 198403 by Mingma last updated on 19/Oct/23 Answered by MM42 last updated on 19/Oct/23 $$\sqrt{{x}+\mathrm{2}}={u}\Rightarrow{x}={u}^{\mathrm{2}} −\mathrm{2}\Rightarrow{dx}=\mathrm{2}{udu} \\ $$$$\Rightarrow\int\:\frac{\mathrm{2}{udu}}{{u}^{\mathrm{2}} −{u}−\mathrm{2}}=\frac{\mathrm{2}}{\mathrm{3}}\int\left(\frac{\mathrm{2}}{{u}−\mathrm{2}}+\frac{\mathrm{1}}{{u}+\mathrm{1}}\right){du} \\ $$$$=\frac{\mathrm{4}}{\mathrm{3}}{ln}\left({u}−\mathrm{2}\right)+\frac{\mathrm{2}}{\mathrm{3}}{ln}\left({u}+\mathrm{1}\right)+{c} \\ $$$$=\frac{\mathrm{4}}{\mathrm{3}}{ln}\left(\sqrt{{x}+\mathrm{2}}−\mathrm{2}\right)+\frac{\mathrm{2}}{\mathrm{3}}{ln}\left(\sqrt{{x}+\mathrm{2}}+\mathrm{1}\right)+{c}\:\:\checkmark…
Question Number 198001 by mathlove last updated on 07/Oct/23 $$\int\left({e}\right)^{\left({x}\right)^{{lnx}} } \:{dx}=? \\ $$ Commented by Frix last updated on 07/Oct/23 $$\left(\mathrm{e}\right)^{\left({x}\right)^{\mathrm{ln}\:{x}} } =\mathrm{e}^{\left({x}^{\mathrm{ln}\:{x}} \right)}…
Question Number 197919 by universe last updated on 04/Oct/23 $$\:\:\:\:\:\:\mathrm{I}_{{m}} \:\:\:\:\:=\:\:\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \left(\frac{\lfloor\mathrm{2}^{{m}} {x}\rfloor}{\mathrm{3}^{{m}} }\:\underset{{n}={m}+\mathrm{1}} {\overset{\infty} {\sum}}\frac{\lfloor\mathrm{2}^{{n}} {x}\rfloor}{\mathrm{3}^{{n}} }\right){dx} \\ $$$$\:\:\:\:\:\:\mathrm{then}\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of} \\ $$$$\:\:\:\:\:\:\mathrm{I}\:=\:\:\:\underset{{m}=\mathrm{1}} {\overset{\infty} {\sum}}\mathrm{I}_{{m}}…
Question Number 197906 by mnjuly1970 last updated on 03/Oct/23 $$ \\ $$$$\:\:\:\:\:\:\:\mathrm{S}=\:\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\:\Gamma^{\:\mathrm{2}} \left(\:{k}\:\right)}{{k}\:\Gamma\:\left(\mathrm{2}{k}\:\right)}\:=\:? \\ $$$$\:\:\:\:\:\:\:\:−−−− \\ $$ Answered by Dwan last updated on…
Question Number 197821 by mnjuly1970 last updated on 30/Sep/23 $$ \\ $$$$\:\:\:\:{find}\:{the}\:{value}\:\:{of}\:: \\ $$$$ \\ $$$$\:\:\:\:\:\Omega\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:\:\mathrm{ln}\:\left(\:\mathrm{1}+\:\frac{\mathrm{1}}{{x}^{\:\mathrm{2}} }\:\right)}{\mathrm{2}\:+\:{x}^{\:\mathrm{2}} }\:{dx}\:=\:? \\ $$$$ \\ $$ Answered…