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Question-221583

Question Number 221583 by MrGaster last updated on 08/Jun/25 Answered by MrGaster last updated on 08/Jun/25 $$=\int_{\mathrm{0}} ^{\mathrm{1}} {t}^{\mathrm{1}/\mathrm{11}−\mathrm{1}} \left(\mathrm{1}−{t}\right)^{\mathrm{9}/\mathrm{11}−\mathrm{1}} \beta\left(\frac{\:\mathrm{1}}{\mathrm{11}},\frac{\mathrm{9}}{\mathrm{11}}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{11}}\right)\Gamma\left(\frac{\mathrm{9}}{\mathrm{11}}\right)\left(\int_{\mathrm{0}} ^{\mathrm{1}} {s}^{\mathrm{3}/\mathrm{11}−\mathrm{1}} \left(\mathrm{1}−{s}\right)^{\mathrm{5}/\mathrm{11}−\mathrm{1}} \beta\left(\frac{\mathrm{3}}{\mathrm{11}},\frac{\mathrm{5}}{\mathrm{11}}\right)\Gamma\left(\frac{\mathrm{3}}{\mathrm{11}}\right)\Gamma\left(\frac{\mathrm{5}}{\mathrm{11}}\right)\int_{\mathrm{0}}…

Prove-0-1-xdx-x-2-1-e-2pix-1-2-1-4-where-is-a-Euler-s-Mascheroni-constant-

Question Number 221577 by Nicholas666 last updated on 08/Jun/25 $$ \\ $$$$\:\:\:\:\:\:\mathrm{Prove};\:\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\frac{{x}\mathrm{d}{x}}{\left({x}^{\mathrm{2}} \:+\:\mathrm{1}\right)\left({e}^{\mathrm{2}\pi{x}} \:−\:\mathrm{1}\right)}\:=\:\frac{\gamma}{\mathrm{2}}\:−\:\frac{\mathrm{1}}{\mathrm{4}}\:\: \\ $$$$\:\:\:\:\mathrm{where};\:\gamma\:\:\mathrm{is}\:\mathrm{a}\:\mathrm{Euler}'\mathrm{s}\:\mathrm{Mascheroni}\:\mathrm{constant}\:\:\:\: \\ $$$$ \\ $$ Answered by MrGaster…

sin-2x-1-sin-3x-dx-

Question Number 221601 by Nicholas666 last updated on 08/Jun/25 $$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\int\:\frac{{sin}\:\mathrm{2}{x}}{\mathrm{1}\:+\:{sin}\:\mathrm{3}{x}}\:{dx} \\ $$$$ \\ $$ Answered by Frix last updated on 08/Jun/25 $$=−\mathrm{2}\int\frac{\mathrm{cos}\:{x}\:\mathrm{sin}\:{x}}{\mathrm{4sin}^{\mathrm{3}} \:{x}\:−\mathrm{3sin}\:{x}\:+\mathrm{1}}{dx}\:\overset{\left[{t}=\mathrm{sin}\:{x}\right]}…

Prove-0-1-x-ln-x-2-1-x-2-2-4x-2-2x-4-3x-6-2x-8-x-10-dx-21-1024-3-pi-3-1024-pi-3-324-3

Question Number 221469 by Nicholas666 last updated on 06/Jun/25 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{Prove}; \\ $$$$\:\:\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\frac{\left(\sqrt{{x}\:}\:\mathrm{ln}\:\sqrt{{x}}\right)^{\mathrm{2}} }{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{2}} \:+\:\mathrm{4}{x}^{\mathrm{2}} \:+\:\mathrm{2}{x}^{\mathrm{4}} \:+\:\mathrm{3}{x}^{\mathrm{6}} \:+\:\mathrm{2}{x}^{\mathrm{8}} \:+\:{x}^{\mathrm{10}} }\:\:\mathrm{d}{x} \\ $$$$\:\:\:\:\:\:\:=\:\frac{\mathrm{21}}{\mathrm{1024}}\:\zeta\left(\mathrm{3}\right)\:−\:\frac{\pi^{\mathrm{3}} }{\mathrm{1024}}\:+\:\frac{\pi^{\mathrm{3}}…

I-0-1-4-1-1-x-2-y-2-x-2-z-2-x-2-w-2-y-2-z-2-y-2-w-2-z-2-w-2-dxdydzdw-

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}}…

Prove-1-x-1-3x-2-3x-4-2x-6-ln-x-1-2-ln-x-ln-x-1-dx-ln-3-4-

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…

I-0-1-3-ln-1-xy-yz-zx-1-xyz-dxdydz-for-0-1-

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…