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Question Number 93908 by abdomathmax last updated on 16/May/20 | ||
$${calculste}\:\:{lim}_{{x}\rightarrow\mathrm{0}} \:\:\:\:\:\int_{{x}} ^{\mathrm{2}{x}} \:\frac{{arctan}\left({xt}\right)}{{t}+{x}}{dt} \\ $$ | ||
Commented by redmiiuser last updated on 16/May/20 | ||
$${Answer}:\mathrm{0} \\ $$ | ||
Commented by abdomathmax last updated on 17/May/20 | ||
$${let}\:{F}\left({x}\right)=\int_{{x}} ^{\mathrm{2}{x}} \:\frac{{arctan}\left({xt}\right)}{{t}+{x}}\:{dt}\:{we}\:{do}\:{the}\:{changement} \\ $$$${xt}\:={u}\:\Rightarrow{F}\left({x}\right)\:=\int_{{x}^{\mathrm{2}} } ^{\mathrm{2}{x}^{\mathrm{2}} } \:\:\frac{{arctan}\left({u}\right)}{\frac{{u}}{{x}}\:+{x}}\:\frac{{du}}{{x}} \\ $$$$\left.=\int_{{x}^{\mathrm{2}} } ^{\mathrm{2}{x}^{\mathrm{2}} } \:\:\:\frac{{arctanu}}{{u}+{x}^{\mathrm{2}} }{du}\:\:\:\:\exists\:{c}_{{x}} \in\right]{x}^{\mathrm{2}} \:,\mathrm{2}{x}^{\mathrm{2}} \left[\:/\right. \\ $$$${F}\left({x}\right)\:={arctsn}\left({c}_{{x}} \right)\int_{{x}^{\mathrm{2}} } ^{\mathrm{2}{x}^{\mathrm{2}} } \:\frac{{du}}{{u}+{x}^{\mathrm{2}} }\:={arctan}\left({c}_{{x}} \right)\left[{ln}\left({u}+{x}^{\mathrm{2}} \right)\right]_{{x}^{\mathrm{2}} } ^{\mathrm{2}{x}^{\mathrm{2}} } \\ $$$$={arctan}\left({c}_{{x}} \right)×\left({ln}\left(\frac{\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{2}{x}^{\mathrm{2}} }\right)\right)\:\:{but}\:{x}\rightarrow\mathrm{0}\:\Rightarrow{c}_{{x}} \rightarrow\mathrm{0}\:\Rightarrow \\ $$$${lim}_{{x}\rightarrow\mathrm{0}} \:\:{F}\left({x}\right)\:=\mathrm{0}×{ln}\left(\frac{\mathrm{3}}{\mathrm{2}}\right)\:=\mathrm{0} \\ $$ | ||