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0-e-pix-e-x-x-e-pix-1-e-x-1-dx-

Question Number 101828 by floor(10²Eta[1]) last updated on 05/Jul/20 $$\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{\mathrm{e}^{\pi\mathrm{x}} −\mathrm{e}^{\mathrm{x}} }{\mathrm{x}\left(\mathrm{e}^{\pi\mathrm{x}} +\mathrm{1}\right)\left(\mathrm{e}^{\mathrm{x}} +\mathrm{1}\right)}\mathrm{dx} \\ $$ Answered by maths mind last updated on…

dx-1-x-1-x-

Question Number 167360 by LEKOUMA last updated on 14/Mar/22 $$\int\frac{{dx}}{\mathrm{1}+\sqrt{{x}}+\sqrt{\mathrm{1}+{x}}} \\ $$ Answered by MJS_new last updated on 14/Mar/22 $$\int\frac{{dx}}{\mathrm{1}+\sqrt{{x}}+\sqrt{{x}+\mathrm{1}}}= \\ $$$$\:\:\:\:\:\left[{t}=\sqrt{{x}}+\sqrt{{x}+\mathrm{1}}\:\rightarrow\:{dx}=\frac{\left({t}−\mathrm{1}\right)\left({t}+\mathrm{1}\right)\left({t}^{\mathrm{2}} +\mathrm{1}\right)}{\mathrm{2}{t}^{\mathrm{3}} }{dt}\right] \\…

Question-101778

Question Number 101778 by mathocean1 last updated on 04/Jul/20 Answered by mr W last updated on 04/Jul/20 $${f}\left({x}\right)={P}\left({x}\right) \\ $$$$\frac{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{1}}{{x}+\mathrm{1}}={x}^{\mathrm{2}} \\ $$$${x}^{\mathrm{3}} −\mathrm{4}{x}−\mathrm{1}=\mathrm{0} \\…

solve-3-x-2-4-x-6-6-

Question Number 167282 by abdurehime last updated on 11/Mar/22 $$\mathrm{solve}\:\left(\mathrm{3}^{\mathrm{x}} \right)^{\mathrm{2}} ×\mathrm{4}^{\mathrm{x}} =\mathrm{6}\sqrt{\mathrm{6}} \\ $$ Answered by nurtani last updated on 11/Mar/22 $$\left(\mathrm{3}^{{x}} \right)^{\mathrm{2}} ×\mathrm{4}^{{x}}…

Version-2-091-is-available-Slightly-darker-characters-are-used-by-default-A-preference-setting-is-available-to-revert-to-previous-font-Change-setting-and-restart-app-A-new-menu

Question Number 101730 by Tinku Tara last updated on 04/Jul/20 $$\mathrm{Version}\:\mathrm{2}.\mathrm{091}\:\mathrm{is}\:\mathrm{available}: \\ $$$$-\:\mathrm{Slightly}\:\mathrm{darker}\:\mathrm{characters}\:\mathrm{are} \\ $$$$\:\:\:\:\mathrm{used}\:\mathrm{by}\:\mathrm{default}. \\ $$$$\:\:\:\:\mathrm{A}\:\mathrm{preference}\:\mathrm{setting}\:\mathrm{is}\:\mathrm{available} \\ $$$$\:\:\:\:\mathrm{to}\:\mathrm{revert}\:\mathrm{to}\:\mathrm{previous}\:\mathrm{font}. \\ $$$$\:\:\:\:\mathrm{Change}\:\mathrm{setting}\:\mathrm{and}\:\mathrm{restart}\:\mathrm{app}. \\ $$$$-\:\mathrm{A}\:\mathrm{new}\:\mathrm{menu}\:\mathrm{option}\:\mathrm{mark}\:\mathrm{as} \\ $$$$\:\:\:\mathrm{answered}\:\mathrm{is}\:\mathrm{added}.\:\mathrm{This}\:\mathrm{just}\:\mathrm{mark}…

Question-167252

Question Number 167252 by mkam last updated on 10/Mar/22 Answered by mr W last updated on 11/Mar/22 $$\frac{\partial{f}}{\partial{t}}=−\mathrm{2}{x}×{r}\:\mathrm{sin}\:{t}+\mathrm{2}{x}×{r}\:\mathrm{cos}\:{t}−\mathrm{2}{y}\:{r}\:\mathrm{sin}\:{t}+\mathrm{2}{y}×{r}\:\mathrm{cos}\:{t} \\ $$$$\frac{\partial{f}}{\partial{r}}=\mathrm{2}{x}×\mathrm{cos}\:{t}+\mathrm{2}{x}×\mathrm{sin}\:{t}+\mathrm{2}{y}\:\mathrm{cos}\:{t}+\mathrm{2}{y}×\mathrm{sin}\:{t} \\ $$ Terms of Service…