Category: Relation and Functions

Question Number 145746 by mathmax by abdo last updated on 07/Jul/21 $$\mathrm{find}{\int }_0^1\log \left(\mathrm{x}\right)\log (1-\mathrm{x})\log (1-{\mathrm{x}}^2)\mathrm{dx}$$ Terms of Service Privacy Policy Contact: info@tinkutara.com

Question Number 80180 by mathocean1 last updated on 31/Jan/20 $$\mathrm{h}\left(x\right)=\frac{x-x^2}{x+1}$$$$wedefinedthisfunctionon$$$$\mathbb{R}-\{-1\}\to \mathbb{R}$$$$$$$$1)\mathrm{Study}\mathrm{the}\mathrm{variations}\mathrm{of}\mathrm{h}\mathrm{then}$$$$\mathrm{draw}\mathrm{up}\mathrm{its}\mathrm{table}\mathrm{of}\mathrm{variation}.$$$$$$$$\mathrm{please}\:\mathrm{sirs}\:\mathrm{i}\:\mathrm{need}\:\mathrm{your}\:\mathrm{kind}\:\mathrm{help}…

Question Number 145634 by mathmax by abdo last updated on 06/Jul/21 $$\mathrm{let}\mathrm{s}\left(\mathrm{x}\right)=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{{({2\mathrm{x}}^2+2\mathrm{x}\sqrt{1+{\mathrm{x}}^2}+1)}^{\mathrm{n}}}$$$$1)\mathrm{explicite}\mathrm{s}\left(\mathrm{x}\right)$$$$\left.\mathrm{2}\right)\:\mathrm{calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{s}\left(\mathrm{x}\right)\mathrm{dx} \…

Question Number 145514 by mathmax by abdo last updated on 05/Jul/21 $$\mathrm{let}{\mathrm{A}}_{\mathrm{n}}={\int }_0^{2\mathrm{n}\pi }\frac{\mathrm{dx}}{{(2+\mathrm{cosx})}^2}$$$$\mathrm{explicit}{\mathrm{A}}_{\mathrm{n}}\mathrm{and}\mathrm{determine}\mathrm{nature}\mathrm{of}\mathrm{serie}\mathrm{\Sigma }{\mathrm{A}}_{\mathrm{n}}$$ Answered by mathmax by…

Question Number 14353 by tawa tawa last updated on 30/May/17 $$\mathrm{For}\mathrm{all}\mathrm{a},\mathrm{b}\in \mathrm{G},\mathrm{where}\mathrm{G}\mathrm{is}\mathrm{a}\mathrm{group}\mathrm{with}\mathrm{respect}\mathrm{to}\mathrm{operation}{}^{\prime }{\mathrm{o}}^{\prime }$$$$\mathrm{Prove}\mathrm{that}{\left(\mathrm{aob}\right)}^{-1}={\mathrm{b}}^{-1}\mathrm{o}{\mathrm{a}}^{-1}$$ Commented by Tinkutara last updated on 31/May/17 $$\mathrm{Sorry},\:\mathrm{I}\:\mathrm{want}\:\mathrm{to}\:\mathrm{say}\:\mathrm{that}\:\mathrm{are}\:{a}\:\mathrm{and}\:{b}…