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Author: Tinku Tara

1-x-a-3-x-a-3-dx-

Question Number 201110 by emilagazade last updated on 29/Nov/23 $$\int\frac{\mathrm{1}}{\:\sqrt{\left({x}−{a}\right)^{\mathrm{3}} }+\sqrt{\left({x}+{a}\right)^{\mathrm{3}} }}{dx} \\ $$ Answered by Frix last updated on 29/Nov/23 $$\sqrt{{p}}+\sqrt{{q}}=\sqrt{{p}+\mathrm{2}\sqrt{{pq}}+{q}} \\ $$$$\int\frac{{dx}}{\:\left({x}−{a}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} +\left({x}+{a}\right)^{\frac{\mathrm{3}}{\mathrm{2}}}…

two-weels-those-have-the-same-materials-with-radii-r-1-4-and-r-2-14-are-starting-to-move-on-a-surface-with-the-same-velocity-from-x-0-to-x-20-the-surface-has-no-friction-wich-one-arrives-faster-

Question Number 201107 by behi834171 last updated on 29/Nov/23 $${two}\:{weels},\:{those}\:{have}\:{the}\:{same}\:{materials}, \\ $$$${with}\:{radii}:\boldsymbol{{r}}_{\mathrm{1}} =\mathrm{4}\:{and}\:\boldsymbol{{r}}_{\mathrm{2}} =\mathrm{14} \\ $$$${are}\:{starting}\:{to}\:{move}\:{on}\:{a}\:{surface},{with} \\ $$$${the}\:{same}\:{velocity},{from}:\boldsymbol{{x}}=\mathrm{0}\:{to}\:\boldsymbol{{x}}=\mathrm{20}. \\ $$$${the}\:{surface}\:{has}\:{no}\:{friction}. \\ $$$${wich}\:{one}\:{arrives}\:{faster}? \\ $$$${any}\:{informations}\:{needed}? \\…

Question-201091

Question Number 201091 by MrGHK last updated on 29/Nov/23 Commented by Frix last updated on 29/Nov/23 $$\mathrm{Look}\:\mathrm{at}\:\mathrm{the}\:\mathrm{first}\:\mathrm{few}\:\mathrm{summands}: \\ $$$${i}=\mathrm{0}\:\rightarrow\:\mathrm{1} \\ $$$${i}=\mathrm{1}\:\rightarrow\:\frac{{n}}{\mathrm{2}{n}+\mathrm{4}} \\ $$$${i}=\mathrm{2}\:\rightarrow\:\frac{{n}^{\mathrm{2}} −{n}}{\mathrm{6}{n}+\mathrm{24}{n}+\mathrm{24}} \\…

Prove-that-0-2arctan-t-x-e-2-t-1-dt-In-x-xIn-x-x-1-2-In-2-x-Michael-faraday-

Question Number 201011 by Calculusboy last updated on 28/Nov/23 $$\boldsymbol{{Prove}}\:\boldsymbol{{that}} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{2}\boldsymbol{{arctan}}\left(\frac{\boldsymbol{{t}}}{\boldsymbol{{x}}}\right)}{\boldsymbol{{e}}^{\mathrm{2}\boldsymbol{\pi{t}}} −\mathrm{1}}\boldsymbol{{dt}}=\boldsymbol{{In}\Gamma}\left(\boldsymbol{{x}}\right)−\boldsymbol{{xIn}}\left(\boldsymbol{{x}}\right)+\boldsymbol{{x}}−\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{{In}}\left(\frac{\mathrm{2}\boldsymbol{\pi}}{\boldsymbol{{x}}}\right) \\ $$$$\boldsymbol{{Michael}}\:\boldsymbol{{faraday}} \\ $$ Terms of Service Privacy Policy Contact:…