Question Number 67020 by mathmax by abdo last updated on 21/Aug/19
$${find}\:{f}\left({x}\right)\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{arctan}\left(\mathrm{1}+{xt}\right){dt}\:\:{with}\:{x}\:{real} \\ $$
Commented by mathmax by abdo last updated on 22/Aug/19
$${f}\left({x}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \:{arctan}\left(\mathrm{1}+{xt}\right){dt}\:\:{by}\:{parts}\: \\ $$$$\left.{f}\left({x}\right)\:={t}\:{arctan}\left(\mathrm{1}+{xt}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \:−\int_{\mathrm{0}} ^{\mathrm{1}} \:{t}\:\frac{{x}}{\mathrm{1}+\left(\mathrm{1}+{xt}\right)^{\mathrm{2}} }{dt} \\ $$$$={arctan}\left(\mathrm{1}+{x}\right)−{x}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{t}}{\mathrm{1}+\left(\mathrm{1}+{xt}\right)^{\mathrm{2}} }{dt}\:{but} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{t}}{\mathrm{1}+\left(\mathrm{1}+{xt}\right)^{\mathrm{2}} }{dt}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{tdt}}{\mathrm{1}+\mathrm{1}+\mathrm{2}{xt}\:+{x}^{\mathrm{2}} {t}^{\mathrm{2}} }\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{tdt}}{{x}^{\mathrm{2}} {t}^{\mathrm{2}} \:+\mathrm{2}{xt}\:+\mathrm{2}} \\ $$$${let}\:{decompose}\:{F}\left({t}\right)=\frac{{t}}{{x}^{\mathrm{2}} {t}^{\mathrm{2}} \:+\mathrm{2}{xt}+\mathrm{2}} \\ $$$$\Delta^{'} \:={x}^{\mathrm{2}} −\mathrm{2}{x}^{\mathrm{2}} \:=−{x}^{\mathrm{2}} <\mathrm{0}\:\:{if}\:{x}\neq\mathrm{0}\:{so}\: \\ $$$${F}\left({t}\right)\:=\frac{{t}}{{x}^{\mathrm{2}} \left({t}^{\mathrm{2}} \:+\frac{\mathrm{2}{t}}{{x}}+\frac{\mathrm{2}}{{x}^{\mathrm{2}} }\right)}\:=\frac{{t}}{{x}^{\mathrm{2}} \left({t}^{\mathrm{2}} \:+\frac{\mathrm{2}{t}}{{x}}\:+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:+\frac{\mathrm{2}}{{x}^{\mathrm{2}} }−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)} \\ $$$$=\frac{{t}}{{x}^{\mathrm{2}} \left\{\left({t}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right\}}\:\:\:{we}\:{use}\:{the}\:{changement}\:{t}+\frac{\mathrm{1}}{{x}}\:=\frac{\mathrm{1}}{\mid{x}\mid}{u}\:\Rightarrow{xt}+\mathrm{1}={s}\left({x}\right){u} \\ $$$$\int_{{s}\left({x}\right)} ^{\left({x}+\mathrm{1}\right){s}\left({x}\right)} \:\:\frac{{t}}{\mathrm{1}+\left(\mathrm{1}+{tx}\right)^{\mathrm{2}} }{dt}\:=\int_{{s}\left({x}\right)} ^{\left({x}+\mathrm{1}\right){s}\left({x}\right)} \:\:\frac{\left(\frac{\mathrm{1}}{\mid{x}\mid}{u}−\frac{\mathrm{1}}{{x}}\right)}{{x}^{\mathrm{2}} ×\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\left(\mathrm{1}+{u}^{\mathrm{2}} \right)}{du} \\ $$$$=\frac{\mathrm{1}}{\mid{x}\mid}\int_{{s}\left({x}\right)} ^{\left({x}+\mathrm{1}\right){s}\left({x}\right)} \:\:\frac{{u}}{\mathrm{1}+{u}^{\mathrm{2}} }−\frac{\mathrm{1}}{{x}}\:\int_{{s}\left({x}\right)} ^{\left({x}+\mathrm{1}\right){s}\left({x}\right)} \:\frac{{du}}{\mathrm{1}+{u}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\mid{x}\mid}\left[{ln}\left(\mathrm{1}+{u}^{\mathrm{2}} \right)\right]_{{s}\left({x}\right)} ^{\left({x}+\mathrm{1}\right){s}\left({x}\right)} \:−\frac{\mathrm{1}}{{x}}\left[{arctanu}\right]_{{s}\left({x}\right)} ^{\left({x}+\mathrm{1}\right){s}\left({x}\right)} \:\:\:\:\:\left({s}^{\mathrm{2}} \left({x}\right)=\mathrm{1}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\mid{x}\mid}\left\{{ln}\left(\mathrm{1}+\left({x}+\mathrm{1}\right)^{\mathrm{2}} \right)−{ln}\mathrm{2}\right\}−\frac{\mathrm{1}}{{x}}\left\{{arctan}\left({x}+\mathrm{1}\right){s}\left({x}\right)−{arctan}\left({s}\left({x}\right)\right)\right\} \\ $$$$\Rightarrow{f}\left({x}\right)=\:{arctan}\left(\mathrm{1}+{x}\right)−\frac{{s}\left({x}\right)}{\mathrm{2}}\left\{{ln}\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{2}\right)−{ln}\left(\mathrm{2}\right)\right\} \\ $$$$+{arctan}\left({x}+\mathrm{1}\right){s}\left({x}\right)+{x}\:{arctan}\left({s}\left({x}\right)\right) \\ $$$${with}\:{s}\left({x}\right)\:=\mathrm{1}\:{if}\:{x}>\mathrm{0}\:{and}\:{s}\left({x}\right)=−\mathrm{1}\:{if}\:{x}<\mathrm{0} \\ $$