Question Number 221484 by Nicholas666 last updated on 06/Jun/25 $$ \\ $$$$\:\:\:{I}\:=\:\int_{\:\left[\mathrm{0},\mathrm{1}\right]^{\:\mathrm{4}} } \:\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} {y}^{\mathrm{2}} −{x}^{\mathrm{2}} {z}^{\mathrm{2}} −{x}^{\mathrm{2}} {w}^{\mathrm{2}} −{y}^{\mathrm{2}} {z}^{\mathrm{2}} −{y}^{\mathrm{2}} {w}^{\mathrm{2}} −{z}^{\mathrm{2}} {w}^{\mathrm{2}}…
Question Number 221453 by gregori last updated on 06/Jun/25 Answered by Rasheed.Sindhi last updated on 06/Jun/25 $${x}^{\mathrm{3}} =−{x}^{\mathrm{2}} −{x}−\mathrm{1}\left({mod}\:{x}^{\mathrm{3}} +{x}^{\mathrm{2}} +{x}+\mathrm{1}\right) \\ $$$${x}^{\mathrm{4}} \equiv−{x}^{\mathrm{3}} −{x}^{\mathrm{2}}…
Question Number 221454 by Nicholas666 last updated on 06/Jun/25 $$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{Prove} \\ $$$$\:\:\underset{\:\mathrm{1}} {\int}^{\:+\infty} \:\frac{{x}}{\left(−\mathrm{1}\:+\:\mathrm{3}{x}^{\mathrm{2}} \:−\:\mathrm{3}{x}^{\mathrm{4}} \:+\:\mathrm{2}{x}^{\mathrm{6}} \right)\left(\mathrm{ln}\left({x}−\mathrm{1}\right)\:−\:\mathrm{2}\:\mathrm{ln}\:{x}\:+\:\mathrm{ln}\left({x}+\mathrm{1}\right)\right)}\:\:\mathrm{d}{x}\:=\:−\:\frac{\mathrm{ln}\:\mathrm{3}}{\mathrm{4}}\:\:\:\:\:\:\:\:\: \\ $$$$ \\ $$ Answered by…
Question Number 221481 by MrGaster last updated on 06/Jun/25 $$\mathrm{prove}:\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\frac{{k}!\left({n}−{k}\right)!}{{n}!}=−\frac{{i}\mathrm{2}^{{n}−\mathrm{1}} \Gamma\left({n}+\mathrm{2}\right)\left(\pi−{iB}_{\mathrm{2}} \left({n}+\mathrm{2},\mathrm{0}\right)\right)}{{n}!} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 221483 by Nicholas666 last updated on 06/Jun/25 $$ \\ $$$$\:\:\:\:{I}\left(\alpha\right)\:=\:\int\int\int_{\:\left[\mathrm{0},\mathrm{1}\right]^{\mathrm{3}} } \:\frac{\mathrm{ln}\left(\mathrm{1}\:+\:\alpha\left({xy}\:+\:{yz}\:+\:{zx}\right)\right)}{\mathrm{1}\:−\:{xyz}\:}\:{dxdydz}\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{for}\:\alpha\:\in\:\left(\mathrm{0},\mathrm{1}\right) \\ $$$$ \\ $$ Commented by MrGaster last updated…
Question Number 221473 by wewji12 last updated on 06/Jun/25 Answered by MathematicalUser2357 last updated on 06/Jun/25 $$\mathrm{279}\pi \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 221438 by Nicholas666 last updated on 05/Jun/25 $$ \\ $$$$\:\:\:\int_{\:\mathrm{1}} ^{\infty} \frac{\mathrm{ln}\left(\mathrm{ln}\:{x}\right)}{\mathrm{1}−\mathrm{2}{x}\:\mathrm{cos}\:\theta\:+\:{x}^{\mathrm{2}} }\:{dx}\:;\:\mathrm{for}\:\mathrm{all}\:\theta\:\in\:\left(−\pi\:,\:\pi\right)\:\:\:\:\: \\ $$$$ \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 221435 by wewji12 last updated on 05/Jun/25 $$\mathrm{g};\mathbb{R}\rightarrow\mathbb{R}\:,\:\mathrm{g}\in\mathcal{C}^{\omega} \:\mathrm{at}\:\mathbb{R}\:\mathrm{space} \\ $$$$\:\mathrm{evauate}\: \\ $$$$−\int_{\mathrm{0}} ^{\infty} \int_{\mathrm{0}} ^{\infty} \:{y}\centerdot\mathrm{g}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)\mathrm{d}{x}\mathrm{d}{y} \\ $$$$\mathrm{when}\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{z}^{\mathrm{2}}…
Question Number 221444 by Tawa11 last updated on 05/Jun/25 Commented by Tawa11 last updated on 05/Jun/25 In the ∆ABC |AB|=|BC|=|AC|=K, Area of the shaded ∆…
Question Number 221430 by pronation last updated on 05/Jun/25 Answered by wewji12 last updated on 05/Jun/25 $$\mathrm{Automorphism}\:\mathrm{group}..?? \\ $$ Terms of Service Privacy Policy Contact:…