Question Number 213215 by Spillover last updated on 01/Nov/24 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 213216 by Spillover last updated on 01/Nov/24 Answered by MrGaster last updated on 01/Nov/24 $$=\int_{\mathrm{0}} ^{\infty} \frac{{x}^{\mathrm{3}} }{\left({x}+\mathrm{1}\right)^{\mathrm{2}} \left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}^{\mathrm{3}} +\mathrm{1}\right)}{dx}+\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{log}^{\mathrm{2}}…
Question Number 213217 by Spillover last updated on 01/Nov/24 Answered by MrGaster last updated on 01/Nov/24 $$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\begin{pmatrix}{\mathrm{2}{n}}\\{{n}}\end{pmatrix}+{n}}{\mathrm{4}^{{n}} \left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\begin{pmatrix}{\mathrm{2}{n}}\\{{n}}\end{pmatrix}}{\mathrm{4}^{{n}} \left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }+\underset{{n}=\mathrm{1}}…
Question Number 213138 by Spillover last updated on 31/Oct/24 Commented by Frix last updated on 31/Oct/24 $$\mathrm{We}\:\mathrm{must}\:\mathrm{first}\:\mathrm{solve}\:\int\mathrm{sin}\:{t}^{\mathrm{3}} \:{dt}\:\mathrm{which}\:\mathrm{might} \\ $$$$\mathrm{be}\:\mathrm{possible}\:\mathrm{using} \\ $$$$\mathrm{sin}\:{t}^{\mathrm{3}} \:=\frac{\mathrm{e}^{\mathrm{i}{t}^{\mathrm{3}} } −\mathrm{e}^{−\mathrm{i}{t}^{\mathrm{3}}…
Question Number 213139 by Spillover last updated on 31/Oct/24 Answered by MrGaster last updated on 31/Oct/24 $${let}\:{u}=\sqrt{{x}}+\sqrt{{y}}+{z}\Rightarrow{dy}=\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}}}{dx}+\frac{\mathrm{1}}{\mathrm{2}\sqrt{{y}}}{dy}+\frac{\mathrm{1}}{\mathrm{2}\sqrt{{z}}}{dx} \\ $$$$\Rightarrow\mathrm{0}\leq{u}\leq\frac{\mathrm{3}\sqrt{\pi}}{\mathrm{4}} \\ $$$$\Rightarrow{dxdydz}=\mathrm{8}{u}^{\mathrm{2}} {du} \\ $$$$\Leftrightarrow\int_{\mathrm{0}} ^{\frac{\mathrm{3}\sqrt{\pi}}{\mathrm{4}}}…
Question Number 213114 by efronzo1 last updated on 30/Oct/24 $$\:\:\:\:\:\:\:\:\underline{\boldsymbol{\div}} \\ $$ Answered by issac last updated on 30/Oct/24 $$\:\:\:{f}\left({x}\right)=−{C}\left({x}−\mathrm{1}\right)+\frac{\boldsymbol{{i}}}{\pi}\left({x}−\mathrm{1}\right)^{\mathrm{5000}} \mathrm{ln}\left({x}−\mathrm{1}\right) \\ $$$$\left(\mathrm{thx}\:\mathrm{wolfram}\:\mathrm{alpha}!!\right) \\ $$$${f}\left(\mathrm{1}\right)=\mathrm{0}\:,\:\mathrm{cus}\:\:{f}\left(\mathrm{1}+\mathrm{0}\right)+{f}\left(\mathrm{1}−\mathrm{0}\right)=\mathrm{0}^{\mathrm{5000}}…
Question Number 213081 by issac last updated on 30/Oct/24 $$\mathrm{evaluate} \\ $$$$\int_{\mathrm{0}} ^{\:\infty} \:\:\:\:\frac{\mathrm{tanh}\left(\frac{\mathrm{1}}{\mathrm{2}}{z}\right)\mathrm{csch}\left({z}\right)}{{z}}\mathrm{d}{z} \\ $$$$\left.\mathrm{1}\right)\:\mathrm{Complex}\:\mathrm{integral} \\ $$$$\left.\mathrm{2}\right)\:\mathrm{Feynman}\:\mathrm{trick} \\ $$ Answered by Berbere last updated…
Question Number 213098 by MathematicalUser2357 last updated on 30/Oct/24 $$\int\frac{\mathrm{3}{x}+\mathrm{2}}{\mathrm{5}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{3}}\mathrm{d}{x}=? \\ $$ Answered by issac last updated on 30/Oct/24 $$\mathrm{Hmmmmm}….. \\ $$$$\frac{\mathrm{3}{x}+\mathrm{2}}{\mathrm{5}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{3}}=\frac{\mathrm{3}\left(\mathrm{10}{x}+\mathrm{2}\right)}{\mathrm{10}\left(\mathrm{5}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{3}\right)}+\frac{\mathrm{7}}{\mathrm{5}\left(\mathrm{5}{x}^{\mathrm{2}}…
Question Number 213074 by Spillover last updated on 29/Oct/24 Answered by MrGaster last updated on 29/Oct/24 $$=\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \left[−\frac{\mathrm{cos}\left({x}+{y}^{\mathrm{2}} +{z}^{\mathrm{3}} \right)}{\mathrm{3}{z}^{\mathrm{2}} }\right]_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{3}}}…
Question Number 212899 by vasil92 last updated on 26/Oct/24 Answered by MrGaster last updated on 02/Nov/24 $$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\mathrm{1}−{x}^{{n}} }{dx}=\mathrm{1} \\ $$$$=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left[{x}−\frac{{x}^{{n}−\mathrm{1}} }{\left({n}+\mathrm{1}\right)\centerdot^{\mathrm{2}}…