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Question Number 67017 by mathmax by abdo last updated on 21/Aug/19
find ∫ arctan(1+(√(x+1)))dx
$${find}\:\int\:\:{arctan}\left(\mathrm{1}+\sqrt{{x}+\mathrm{1}}\right){dx} \\ $$
Commented by mathmax by abdo last updated on 24/Aug/19
let I =∫ arctan(1+(√(x+1)))dx changement (√(x+1))=t give x+1=t^2 ⇒ I =∫ arctan(1+t)(2t)dt =2 ∫ t arctan(t+1)dt by parts I =2{ (t^2 /2) arctan(t+1)−∫ (t^2 /2)×(1/(1+(t+1)^2 ))dt} =t^2 arctan(t+1)−∫ (t^2 /(1+t^2 +2t+1))dt ∫ (t^2 /(t^(2 ) +2t+2))dt =∫ ((t^2 +2t+2−2t−2)/(t^2 +2t+2))dt =t−∫((2t+2)/(t^2 +2t +2))dt =t−ln(t^2 +2t+2) ⇒I =t^2 arctan(t+1)−t+ln(t^2 +2t+2) +c I=(x+1)arctan(1+(√(x+1)))−(√(x+1)) +ln(x+3+2(√(x+1))) +c
$${let}\:{I}\:=\int\:{arctan}\left(\mathrm{1}+\sqrt{{x}+\mathrm{1}}\right){dx}\:\:{changement}\:\sqrt{{x}+\mathrm{1}}={t}\:{give}\:{x}+\mathrm{1}={t}^{\mathrm{2}} \:\Rightarrow \\ $$$${I}\:=\int\:\:{arctan}\left(\mathrm{1}+{t}\right)\left(\mathrm{2}{t}\right){dt}\:=\mathrm{2}\:\int\:\:{t}\:{arctan}\left({t}+\mathrm{1}\right){dt}\:\:{by}\:{parts} \\ $$$${I}\:=\mathrm{2}\left\{\:\:\:\frac{{t}^{\mathrm{2}} }{\mathrm{2}}\:{arctan}\left({t}+\mathrm{1}\right)−\int\:\:\:\frac{{t}^{\mathrm{2}} }{\mathrm{2}}×\frac{\mathrm{1}}{\mathrm{1}+\left({t}+\mathrm{1}\right)^{\mathrm{2}} }{dt}\right\} \\ $$$$={t}^{\mathrm{2}} \:{arctan}\left({t}+\mathrm{1}\right)−\int\:\:\:\frac{{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} +\mathrm{2}{t}+\mathrm{1}}{dt} \\ $$$$\int\:\:\:\frac{{t}^{\mathrm{2}} }{{t}^{\mathrm{2}\:} +\mathrm{2}{t}+\mathrm{2}}{dt}\:=\int\:\:\frac{{t}^{\mathrm{2}} \:+\mathrm{2}{t}+\mathrm{2}−\mathrm{2}{t}−\mathrm{2}}{{t}^{\mathrm{2}} \:+\mathrm{2}{t}+\mathrm{2}}{dt}\:={t}−\int\frac{\mathrm{2}{t}+\mathrm{2}}{{t}^{\mathrm{2}} \:+\mathrm{2}{t}\:+\mathrm{2}}{dt} \\ $$$$={t}−{ln}\left({t}^{\mathrm{2}} +\mathrm{2}{t}+\mathrm{2}\right)\:\Rightarrow{I}\:={t}^{\mathrm{2}} \:{arctan}\left({t}+\mathrm{1}\right)−{t}+{ln}\left({t}^{\mathrm{2}} +\mathrm{2}{t}+\mathrm{2}\right)\:+{c} \\ $$$${I}=\left({x}+\mathrm{1}\right){arctan}\left(\mathrm{1}+\sqrt{{x}+\mathrm{1}}\right)−\sqrt{{x}+\mathrm{1}}\:+{ln}\left({x}+\mathrm{3}+\mathrm{2}\sqrt{{x}+\mathrm{1}}\right)\:+{c} \\ $$

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