# let-f-x-x-x-2-x-arctan-e-x-t-dt-calculate-f-x-and-f-0-

Question Number 68466 by mathmax by abdo last updated on 11/Sep/19
$${let}\:{f}\left({x}\right)\:=\int_{{x}^{} } ^{{x}^{\mathrm{2}} −{x}} \:{arctan}\left({e}^{−{x}−{t}} \right){dt} \\$$$${calculate}\:{f}^{'} \left({x}\right)\:\:\:{and}\:{f}^{'} \left(\mathrm{0}\right). \\$$
Commented by mathmax by abdo last updated on 11/Sep/19
$${we}\:{have}\:{g}\left({x},{t}\right)={arctan}\left({e}^{−{x}−{t}} \right)\:\:\:,{u}\left({x}\right)={x}\:{and}\:{v}\left({x}\right)={x}^{\mathrm{2}} −{x}\:\Rightarrow \\$$$${f}^{'} \left({x}\right)\:=\int_{{u}\left({x}\right)} ^{{v}\left({x}\right)} \frac{\partial{g}}{\partial{x}}\left({x},{t}\right){dt}\:+{v}^{'} {g}\left({x},{v}\right)−{u}^{'} {g}\left({x},{u}\right) \\$$$$=\int_{{x}} ^{{x}^{\mathrm{2}} −{x}} \:\:\frac{−{e}^{−{x}−{t}} }{\mathrm{1}+{e}^{−\mathrm{2}{x}−\mathrm{2}{t}} }{dt}\:\:+\left(\mathrm{2}{x}−\mathrm{1}\right){g}\left({x},{x}^{\mathrm{2}} −{x}\right)−{g}\left({x},{x}\right) \\$$$$=−{e}^{−{x}} \:\int_{{x}} ^{{x}^{\mathrm{2}} −{x}} \:\:\frac{{e}^{−{t}} }{\mathrm{1}+{e}^{−\mathrm{2}{t}−\mathrm{2}{x}} }{dt}\:\:+\left(\mathrm{2}{x}−\mathrm{1}\right){arctan}\left({e}^{−{x}^{\mathrm{2}} } \right) \\$$$$−{arctan}\left({e}^{−\mathrm{2}{x}} \right)\:\:{let}\:{find}\:{I}\:=\int_{{x}} ^{{x}^{\mathrm{2}} −{x}} \:\frac{{e}^{−{t}} }{\mathrm{1}+{e}^{−\mathrm{2}{t}−\mathrm{2}{x}} }{dt}\:\Rightarrow \\$$$${I}\:=\int_{{x}} ^{{x}^{\mathrm{2}} −\mathrm{1}} \:\:\frac{{e}^{{t}} }{{e}^{\mathrm{2}{t}} +{e}^{−\mathrm{2}{x}} }\:{dt}\:\:\:=_{{e}^{{t}} ={u}} \:\:\:\:\:\int_{{e}^{{x}} } ^{{e}^{{x}^{\mathrm{2}} −{x}} } \:\:\:\:\frac{{u}}{{u}^{\mathrm{2}} \:+{e}^{−\mathrm{2}{x}} }\:\frac{{du}}{{u}} \\$$$$=\int_{{e}^{{x}} } ^{{e}^{{x}^{\mathrm{2}} −{x}} } \:\:\frac{{du}}{{u}^{\mathrm{2}} \:+{e}^{−\mathrm{2}{x}} }\:=_{{u}={e}^{−{x}} \:{z}} \:\:\:\:\int_{{e}^{\mathrm{2}{x}} } ^{{e}^{{x}^{\mathrm{2}} } } \:\:\:\:\frac{{e}^{−{x}} {dz}}{{e}^{−\mathrm{2}{x}} \left({z}^{\mathrm{2}} +\mathrm{1}\right)}\:={e}^{{x}} \left[{arctanz}\right]_{{e}^{\mathrm{2}{x}} } ^{{e}^{{x}^{\mathrm{2}} } } \\$$$$={e}^{{x}} \left\{\:{arctan}\left({e}^{{x}^{\mathrm{2}} } \right)−{arctan}\left({e}^{\mathrm{2}{x}} \right)\right\}\:\Rightarrow \\$$$${f}^{'} \left({x}\right)\:=−{arctan}\left({e}^{{x}^{\mathrm{2}} } \right)+{e}^{−{x}} \:{arctan}\left({e}^{\mathrm{2}{x}} \right)+\left(\mathrm{2}{x}−\mathrm{1}\right){arctan}\left({e}^{−{x}^{\mathrm{2}} } \right) \\$$$$−{arctan}\left({e}^{−\mathrm{2}{x}} \right)\:. \\$$
Commented by mathmax by abdo last updated on 11/Sep/19
$${f}^{'} \left(\mathrm{0}\right)\:=−\frac{\pi}{\mathrm{4}}+\frac{\pi}{\mathrm{4}}−\frac{\pi}{\mathrm{4}}−\frac{\pi}{\mathrm{4}}\:\Rightarrow{f}^{'} \left(\mathrm{0}\right)\:=−\frac{\pi}{\mathrm{2}} \\$$
Answered by meme last updated on 11/Sep/19
$${f}^{'} \left({x}\right)={arctan}\left({e}^{−{x}−{x}^{\mathrm{2}} +{x}} \right)−{arctan}\left({e}^{−{x}−{x}} \right) \\$$$$\:\:\:\:\:\:\:\:\:\:\:=\:{arctan}\left({e}^{−{x}^{\mathrm{2}} } \right)−{arctan}\left({e}^{−\mathrm{2}{x}} \right) \\$$$$\:\:\:\:{f}^{'} \left(\mathrm{0}\right)={arctan}\left(\mathrm{1}\right)−{arctan}\left(\mathrm{1}\right) \\$$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{0} \\$$
Commented by turbo msup by abdo last updated on 11/Sep/19
$${your}\:{answer}\:{is}\:{not}\:{correct}\:{sir}. \\$$