Question Number 40870 by math khazana by abdo last updated on 28/Jul/18
$${fnd}\:\:\int\:\:\left(\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right){arctan}\left({x}−\frac{\mathrm{1}}{{x}}\right){dx}\:. \\ $$
Commented by maxmathsup by imad last updated on 29/Jul/18
$${changement}\:{x}−\frac{\mathrm{1}}{{x}}={t}\:{give}\:\left(\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right){dx}={dt}\:\Rightarrow \\ $$$${I}\:=\:\int\:\:\:\:{arctant}\:{dt}\:=\:{t}\:{arctant}\:−\int\:\frac{{t}}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}\:+{c} \\ $$$$={t}\:{arctan}\left({t}\right)−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)+{c} \\ $$$$=\left({x}−\frac{\mathrm{1}}{{x}}\right){arctan}\left({x}−\frac{\mathrm{1}}{{x}}\right)−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}+\left({x}−\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} \right)\:+{c} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 28/Jul/18
$${t}={x}−\frac{\mathrm{1}}{{x}}\:\:\:{dt}=\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}\:} }\:{dt} \\ $$$$\int{tan}^{−\mathrm{1}} {dt} \\ $$$${t}\:×{tan}^{−\mathrm{1}} {t}−\int\frac{\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{2}} }×{tdt} \\ $$$${t}×{tan}^{−\mathrm{1}} {t}−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)+{c} \\ $$$$\left({x}−\frac{\mathrm{1}}{{x}}\right){tan}^{−\mathrm{1}} \left({x}−\frac{\mathrm{1}}{{x}}\right)−\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid\left(\mathrm{1}+{x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }−\mathrm{2}\right)\mid+{c} \\ $$$$\left({x}−\frac{\mathrm{1}}{{x}}\right){tan}^{−\mathrm{1}} \left({x}−\frac{\mathrm{1}}{{x}}\right)−\frac{\mathrm{1}}{\mathrm{2}}{ln}\mid{x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }−\mathrm{1}\mid+{c} \\ $$