Question Number 222619 by ahmed2025 last updated on 01/Jul/25 Answered by wewji12 last updated on 02/Jul/25 $$\int\:\:\frac{\mathrm{2}}{\mathrm{2}−\mathrm{3}{x}}\:\:\mathrm{d}{x}=−\frac{\mathrm{1}}{\mathrm{3}}\mathrm{ln}\left(\mathrm{2}−\mathrm{3}{x}\right)+{C} \\ $$$$\int_{−\mathrm{1}} ^{\:\:\mathrm{2}} \:\frac{\mathrm{2}}{\mathrm{2}−\mathrm{3}{x}}\:\mathrm{d}{x}=\int_{\:−\mathrm{1}} ^{\frac{\mathrm{2}}{\mathrm{3}}} \frac{\mathrm{2}}{\mathrm{2}−\mathrm{3}{x}}\:\mathrm{d}{x}\:+\int_{\frac{\mathrm{2}}{\mathrm{3}}} ^{\:\mathrm{2}} \:\frac{\mathrm{2}}{\mathrm{2}−\mathrm{3}{x}}\:\mathrm{d}{x}…
Question Number 222613 by Nicholas666 last updated on 01/Jul/25 $$ \\ $$$$\:\:\:\:\:\:\mathrm{what}\:\mathrm{is}\:\:{I}\:=\:\int\:\mathrm{tan}\:\left(\frac{\mathrm{cos}\:{x}}{{x}}\right)\:\mathrm{d}{x}\: \\ $$$$ \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 222614 by Nicholas666 last updated on 01/Jul/25 $$ \\ $$$$\:\:\mathrm{complicated}\:\mathrm{solution}\:\mathrm{of}\:\mathrm{the}\:\mathrm{integral}\:; \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\int\:\mathrm{tan}\left(\frac{\mathrm{cos}\:{x}}{{x}}\right)^{\mathrm{13}} \:\mathrm{d}{x} \\ $$$$ \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 222599 by MrGaster last updated on 01/Jul/25 $$\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{k}+\mathrm{1}} \frac{\mathrm{1}}{\left(\mathrm{2}{k}−\mathrm{1}\right)^{\mathrm{2}{n}+\mathrm{1}} }=\frac{\pi^{\mathrm{2}{n}+\mathrm{1}} }{\mathrm{2}^{\mathrm{2}{n}+\mathrm{2}} \left(\mathrm{2}{n}\right)!}\mid{E}_{\mathrm{2}{m}} \mid \\ $$ Commented by MathematicalUser2357 last updated on…
Question Number 222610 by Nicholas666 last updated on 01/Jul/25 $$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{{n}\left({e}^{\mathrm{2}\pi{n}} \:−\:\mathrm{1}\right)} \\ $$$$ \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 222595 by Ghisom last updated on 01/Jul/25 $$\Omega=\int\:\frac{\mathrm{sin}^{\mathrm{2}} \:{x}\:\sqrt{\mathrm{cos}\:{x}}}{\mathrm{1}+\mathrm{3sin}^{\mathrm{2}} \:{x}}{dx} \\ $$$$\mathrm{I}\:\mathrm{can}'\mathrm{t}\:\mathrm{solve}\:\mathrm{this},\:\mathrm{maybe}\:\mathrm{some}\:\mathrm{genius}\:\mathrm{can}. \\ $$$$\mathrm{WolframAlpha}\:\mathrm{gives}\:\mathrm{a}\:\mathrm{solution}\:\mathrm{but}\:\mathrm{I} \\ $$$$\mathrm{need}\:\mathrm{the}\:\mathrm{path}… \\ $$ Commented by MrGaster last updated…
Question Number 222584 by hu last updated on 30/Jun/25 $${find}\:{the}\:{correct}\:{const}.\:{to}\:{preconst}.\:“\:{x}^{{n}} −\mathrm{2}=−{x}\:'' \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 222585 by Rojarani last updated on 30/Jun/25 Commented by AntonCWX8 last updated on 01/Jul/25 $${Now}\:{I}'{ll}\:{give}\:{you}\:{guys}\:{another}\:{challenge}. \\ $$$${What}\:{if}\:{the}\:{numbers}\:{beside}\:{the}\:\sqrt{\:\:}\:{symbol}\:{represent}\:{the}\:{nth}\:{root}? \\ $$$${Say}\:\mathrm{5}\sqrt{\:\:\:}\:=\:\sqrt[{\mathrm{5}}]{\:\:\:\:} \\ $$ Commented by…
Question Number 222582 by hardmath last updated on 30/Jun/25 $$\mathrm{If}:\:\:\:\mathrm{a}_{\boldsymbol{\mathrm{i}}} \:>\:\mathrm{0}\:\:\:,\:\:\:\mathrm{b}_{\boldsymbol{\mathrm{i}}} \:>\:\mathrm{0}\:\:\:,\:\:\:\mathrm{i}\:=\:\overline {\mathrm{1},…,\mathrm{n}} \\ $$$$\mathrm{Prove}\:\mathrm{that}: \\ $$$$\sqrt{\mathrm{a}_{\mathrm{1}} ^{\mathrm{2}} \:+\:\mathrm{b}_{\mathrm{1}} ^{\mathrm{2}} }\:+\:\sqrt{\mathrm{a}_{\mathrm{2}} ^{\mathrm{2}} \:+\:\mathrm{b}_{\mathrm{2}} ^{\mathrm{2}} }\:+…+\:\sqrt{\mathrm{a}_{\boldsymbol{\mathrm{n}}}…
Question Number 222583 by ajfour last updated on 30/Jun/25 $${solve}\:{for}\:{p},{q},{s}\:{in}\:{terms}\:{of}\:{c}. \\ $$$$\bullet\:\left(\frac{{qs}}{{q}−{sp}}\right)^{\mathrm{2}} −{s}\left(\frac{{qs}}{{q}−{sp}}\right)+{p}=\mathrm{0} \\ $$$$\bullet\:\left(\frac{{q}+{c}}{{p}+\mathrm{1}}\right)^{\mathrm{2}} ={sp}−{q} \\ $$$$\bullet\:\left({q}−{cp}\right)\left({p}+\mathrm{1}\right)^{\mathrm{2}} =\left({q}+{c}\right)^{\mathrm{3}} \\ $$$${I}\:{have}\:{to}\:{find}\:{non}\:{zero}\:{real}\:{x}=−\left(\frac{{q}+{c}}{{p}+\mathrm{1}}\right)\:. \\ $$ Terms of…