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Question Number 27282 by hp killer last updated on 04/Jan/18

∫log(2+x^2 )dx

$$\int{log}\left(\mathrm{2}+{x}^{\mathrm{2}} \right){dx} \\ $$

Commented by abdo imad last updated on 04/Jan/18

if you mean ln integrate par parts u^′ =1 and v=ln(2+x^2 )  ∫ln(2+x^2 )dx= xln(2 +x^2 ) −∫ x ((2x)/(2+x^2 ))dx  = xln(2+x^2 ) −2 ∫ (x^2 /(2+x^2 ))dx  =xln(2+x^2 ) −2 ∫((2+x^2  −2)/(2+x^2 ))dx  =xln(2+x^2 )−2x  +4 ∫(dx/(2+x^2 ))  and by the changement x=(√2)t  ∫  (dx/(2+x^2 )) = ∫ (((√2)dt)/(2+2t^2 ))= ((√2)/2) artan((x/(√2))) so  ∫ln(2+x^2 )dx= xln(2+x^2 ) −2x + 2(√2) arctan((x/(√2))).

$${if}\:{you}\:{mean}\:{ln}\:{integrate}\:{par}\:{parts}\:{u}^{'} =\mathrm{1}\:{and}\:{v}={ln}\left(\mathrm{2}+{x}^{\mathrm{2}} \right) \\ $$$$\int{ln}\left(\mathrm{2}+{x}^{\mathrm{2}} \right){dx}=\:{xln}\left(\mathrm{2}\:+{x}^{\mathrm{2}} \right)\:−\int\:{x}\:\frac{\mathrm{2}{x}}{\mathrm{2}+{x}^{\mathrm{2}} }{dx} \\ $$$$=\:{xln}\left(\mathrm{2}+{x}^{\mathrm{2}} \right)\:−\mathrm{2}\:\int\:\frac{{x}^{\mathrm{2}} }{\mathrm{2}+{x}^{\mathrm{2}} }{dx} \\ $$$$={xln}\left(\mathrm{2}+{x}^{\mathrm{2}} \right)\:−\mathrm{2}\:\int\frac{\mathrm{2}+{x}^{\mathrm{2}} \:−\mathrm{2}}{\mathrm{2}+{x}^{\mathrm{2}} }{dx} \\ $$$$={xln}\left(\mathrm{2}+{x}^{\mathrm{2}} \right)−\mathrm{2}{x}\:\:+\mathrm{4}\:\int\frac{{dx}}{\mathrm{2}+{x}^{\mathrm{2}} }\:\:{and}\:{by}\:{the}\:{changement}\:{x}=\sqrt{\mathrm{2}}{t} \\ $$$$\int\:\:\frac{{dx}}{\mathrm{2}+{x}^{\mathrm{2}} }\:=\:\int\:\frac{\sqrt{\mathrm{2}}{dt}}{\mathrm{2}+\mathrm{2}{t}^{\mathrm{2}} }=\:\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:{artan}\left(\frac{{x}}{\sqrt{\mathrm{2}}}\right)\:{so} \\ $$$$\int{ln}\left(\mathrm{2}+{x}^{\mathrm{2}} \right){dx}=\:{xln}\left(\mathrm{2}+{x}^{\mathrm{2}} \right)\:−\mathrm{2}{x}\:+\:\mathrm{2}\sqrt{\mathrm{2}}\:{arctan}\left(\frac{{x}}{\sqrt{\mathrm{2}}}\right). \\ $$

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