Question Number 211800 by MrGaster last updated on 21/Sep/24 $$ \\ $$$$\boldsymbol{{set}}\:\boldsymbol{\Omega}=\left\{\left(\boldsymbol{{x}},\boldsymbol{{y}},\boldsymbol{{z}}\right)\mid\boldsymbol{{x}}^{\mathrm{2}} +\boldsymbol{{y}}^{\mathrm{2}} +\boldsymbol{{z}}^{\mathrm{2}} \leq\mathrm{1}\right\}, \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{certificat}}\mathrm{e}: \\ $$$$\:\:\frac{\mathrm{4}\boldsymbol{\pi}}{\mathrm{3}}\sqrt[{\mathrm{3}}]{\mathrm{2}}\leq\int\underset{\boldsymbol{\Omega}} {\int}\int\sqrt[{\mathrm{3}}]{\boldsymbol{{x}}+\mathrm{2}\boldsymbol{{y}}−\mathrm{2}\boldsymbol{{z}}+\mathrm{5}}\boldsymbol{{dv}}\leq\frac{\mathrm{8}\boldsymbol{\pi}}{\mathrm{3}} \\ $$$$ \\ $$$$ \\…
Question Number 211738 by MrGaster last updated on 19/Sep/24 $$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\int\frac{\mathrm{1}}{\boldsymbol{{x}}^{\mathrm{5}} +\mathrm{1}\:}\boldsymbol{{dx}}. \\ $$$$ \\ $$ Commented by Frix last updated on 19/Sep/24 $$=\int\frac{{dx}}{\left({x}+\mathrm{1}\right)\left({x}^{\mathrm{2}}…
Question Number 211703 by MrGaster last updated on 17/Sep/24 $$ \\ $$$$\:\mathrm{Calculate}\:\mathrm{the}\:\mathrm{quadruple}\: \\ $$$$\mathrm{integralas}\:\mathrm{follows}: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{I}}=\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\boldsymbol{{x}}} \int_{\mathrm{0}} ^{\boldsymbol{\mathrm{y}}} \int_{\mathrm{0}} ^{\boldsymbol{{z}}} \frac{\boldsymbol{\mathrm{sin}}\left(\boldsymbol{{x}}^{\mathrm{2}} +\boldsymbol{\mathrm{y}}^{\mathrm{2}}…
Question Number 211579 by MrGaster last updated on 13/Sep/24 $$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\int\frac{\mathrm{1}}{\left(\mathrm{1}−\boldsymbol{{x}}^{\mathrm{4}} \right)\sqrt{\mathrm{1}+\boldsymbol{{x}}^{\mathrm{2}} }}\boldsymbol{{dx}}. \\ $$$$ \\ $$ Answered by lepuissantcedricjunior last updated on 13/Sep/24…
Question Number 211560 by MrGaster last updated on 13/Sep/24 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{certificate}: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{I}}=\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \sqrt{\sqrt{\boldsymbol{{x}}^{\mathrm{2}} +\boldsymbol{\mathrm{ln}}^{\mathrm{2}} \boldsymbol{\mathrm{cos}}\left(\boldsymbol{{x}}\right)−\boldsymbol{\mathrm{lncos}}\left(\boldsymbol{{x}}\right)\boldsymbol{{dx}}}}=\frac{\boldsymbol{\pi}}{\mathrm{2}}\sqrt{\mathrm{2}\boldsymbol{\mathrm{ln}}\mathrm{2}} \\ $$$$ \\ $$ Terms of Service…
Question Number 211546 by MrGaster last updated on 12/Sep/24 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\int\frac{\boldsymbol{{dx}}}{\:\sqrt{\boldsymbol{{x}}^{\mathrm{2}} −\mathrm{4}\boldsymbol{{x}}+\mathrm{13}}}=? \\ $$ Answered by Frix last updated on 12/Sep/24 $$\int\frac{{dx}}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{13}}}\:\overset{\left[{t}=\frac{{x}−\mathrm{2}+\sqrt{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{13}}}{\mathrm{3}}\right]}…
Question Number 210786 by zhou0429 last updated on 19/Aug/24 Answered by Berbere last updated on 19/Aug/24 $${x}^{\mathrm{9}} +\mathrm{1}=\underset{{k}=\mathrm{0}} {\overset{\mathrm{8}} {\prod}}\left({x}−{e}^{{i}\pi\left(\frac{\mathrm{1}+\mathrm{2}{k}}{\mathrm{9}}\right)} \right)={p}\left({x}\right) \\ $$$$=\Sigma\int\frac{\mathrm{1}}{\left({x}−{e}^{{i}\pi\left(\frac{\mathrm{1}+\mathrm{2}{k}}{\mathrm{9}}\right)} \right){p}'\left({e}^{{i}\pi\left(\frac{\mathrm{1}+\mathrm{2}{k}}{\mathrm{9}}\right)} \right)}{dx}…
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Question Number 208569 by pticantor last updated on 18/Jun/24 $$\boldsymbol{{help}}\:\boldsymbol{{me}}\:\boldsymbol{{to}}\:\boldsymbol{{solve}}\:\boldsymbol{{this}}\:\boldsymbol{{please}} \\ $$$$\:\:\boldsymbol{{y}}''−\sqrt{\mathrm{1}+\boldsymbol{{y}}'^{\mathrm{2}} }=\boldsymbol{{x}}^{\mathrm{2}} \\ $$$$\boldsymbol{{solve}}\:\boldsymbol{{this}}\:\boldsymbol{{differential}}\:\boldsymbol{{equation}} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$ Terms of…
Question Number 208303 by Simurdiera last updated on 10/Jun/24 $${Resolver} \\ $$$$\frac{\partial^{\mathrm{2}} {u}}{\partial{y}^{\mathrm{2}} }\:−\:{x}^{\mathrm{2}} {u}\:=\:{xe}^{\mathrm{4}{y}} \\ $$ Answered by aleks041103 last updated on 12/Jun/24 $${u}={X}\left({x}\right){Y}\left({y}\right)…