Question Number 27282 by hp killer last updated on 04/Jan/18
$$\int{log}\left(\mathrm{2}+{x}^{\mathrm{2}} \right){dx} \\ $$
Commented by abdo imad last updated on 04/Jan/18
$${if}\:{you}\:{mean}\:{ln}\:{integrate}\:{par}\:{parts}\:{u}^{'} =\mathrm{1}\:{and}\:{v}={ln}\left(\mathrm{2}+{x}^{\mathrm{2}} \right) \\ $$$$\int{ln}\left(\mathrm{2}+{x}^{\mathrm{2}} \right){dx}=\:{xln}\left(\mathrm{2}\:+{x}^{\mathrm{2}} \right)\:−\int\:{x}\:\frac{\mathrm{2}{x}}{\mathrm{2}+{x}^{\mathrm{2}} }{dx} \\ $$$$=\:{xln}\left(\mathrm{2}+{x}^{\mathrm{2}} \right)\:−\mathrm{2}\:\int\:\frac{{x}^{\mathrm{2}} }{\mathrm{2}+{x}^{\mathrm{2}} }{dx} \\ $$$$={xln}\left(\mathrm{2}+{x}^{\mathrm{2}} \right)\:−\mathrm{2}\:\int\frac{\mathrm{2}+{x}^{\mathrm{2}} \:−\mathrm{2}}{\mathrm{2}+{x}^{\mathrm{2}} }{dx} \\ $$$$={xln}\left(\mathrm{2}+{x}^{\mathrm{2}} \right)−\mathrm{2}{x}\:\:+\mathrm{4}\:\int\frac{{dx}}{\mathrm{2}+{x}^{\mathrm{2}} }\:\:{and}\:{by}\:{the}\:{changement}\:{x}=\sqrt{\mathrm{2}}{t} \\ $$$$\int\:\:\frac{{dx}}{\mathrm{2}+{x}^{\mathrm{2}} }\:=\:\int\:\frac{\sqrt{\mathrm{2}}{dt}}{\mathrm{2}+\mathrm{2}{t}^{\mathrm{2}} }=\:\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:{artan}\left(\frac{{x}}{\:\sqrt{\mathrm{2}}}\right)\:{so} \\ $$$$\int{ln}\left(\mathrm{2}+{x}^{\mathrm{2}} \right){dx}=\:{xln}\left(\mathrm{2}+{x}^{\mathrm{2}} \right)\:−\mathrm{2}{x}\:+\:\mathrm{2}\sqrt{\mathrm{2}}\:{arctan}\left(\frac{{x}}{\:\sqrt{\mathrm{2}}}\right). \\ $$