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Question Number 82510 by jagoll last updated on 22/Feb/20
∫ (√(x+(√x) )) dx = ?
$$\int\:\sqrt{{x}+\sqrt{{x}}\:}\:{dx}\:=\:? \\ $$
Commented by mathmax by abdo last updated on 23/Feb/20
I =∫(√(x+(√x)))dx changement (√x)=t give x=t^2 ⇒ I =∫(√(t^2 +t))(2t)dt =2∫ t(√(t^2 +t))dt =2∫t(√(t^2 +2(1/2)t +(1/(4 ))−(1/4)))dt =2∫ t(√((t+(1/2))^2 −(1/4)))dt =_(t+(1/2)=(1/2)ch(u)) 2∫ (1/2)(ch(u)−1)×(1/2)sh(u)(1/2)sh(u)du =(1/4) ∫ (ch(u)−1)sh^2 (u)du =(1/4)∫ ch(u)sh^2 (u)du −(1/4)∫ ((ch(2u)−1)/2)du =(1/(12))sh^3 (u)−(1/8) ∫ ch(2u)+(1/8)u )+C =(1/(12))sh^2 u shu −(1/(16))sh(2u)+(u/8) +C =(1/(12))((2t+1)^2 −1)(√((2t+1)^2 −1))−(1/8)(2t+1)(√((2t+1)^2 −1))+(1/8)argch(2t+1) +C =(1/(12))(4t^2 +4t)(√(4t^2 +4t))−(1/8)(2t+1)(√(4t^2 +4t))+(1/8)ln(2t+1+(√((2t+1)^2 −1))) +C =(2/3)(t^2 +t)(√(t^2 +t))−(1/4)(2t+1)(√(t^2 +t)) +(1/8)ln(2t+1)+2(√(t^2 +t)))+C ⇒I =(2/3)(x+(√x))(√(x+(√x)))−(1/4)(2(√x)+1)(√(x+(√x)))+(1/8)ln(2(√x)+1+2(√(x+(√x)))) ) +C
$${I}\:=\int\sqrt{{x}+\sqrt{{x}}}{dx}\:{changement}\:\sqrt{{x}}={t}\:{give}\:{x}={t}^{\mathrm{2}} \:\Rightarrow \\ $$$${I}\:=\int\sqrt{{t}^{\mathrm{2}} +{t}}\left(\mathrm{2}{t}\right){dt}\:=\mathrm{2}\int\:{t}\sqrt{{t}^{\mathrm{2}} +{t}}{dt}\:=\mathrm{2}\int{t}\sqrt{{t}^{\mathrm{2}} +\mathrm{2}\frac{\mathrm{1}}{\mathrm{2}}{t}\:+\frac{\mathrm{1}}{\mathrm{4}\:}−\frac{\mathrm{1}}{\mathrm{4}}}{dt} \\ $$$$=\mathrm{2}\int\:{t}\sqrt{\left({t}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}}{dt}\:\:=_{{t}+\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}}{ch}\left({u}\right)} \mathrm{2}\int\:\frac{\mathrm{1}}{\mathrm{2}}\left({ch}\left({u}\right)−\mathrm{1}\right)×\frac{\mathrm{1}}{\mathrm{2}}{sh}\left({u}\right)\frac{\mathrm{1}}{\mathrm{2}}{sh}\left({u}\right){du} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\:\int\:\left({ch}\left({u}\right)−\mathrm{1}\right){sh}^{\mathrm{2}} \left({u}\right){du} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\int\:{ch}\left({u}\right){sh}^{\mathrm{2}} \left({u}\right){du}\:−\frac{\mathrm{1}}{\mathrm{4}}\int\:\frac{{ch}\left(\mathrm{2}{u}\right)−\mathrm{1}}{\mathrm{2}}{du} \\ $$$$\left.=\frac{\mathrm{1}}{\mathrm{12}}{sh}^{\mathrm{3}} \left({u}\right)−\frac{\mathrm{1}}{\mathrm{8}}\:\int\:{ch}\left(\mathrm{2}{u}\right)+\frac{\mathrm{1}}{\mathrm{8}}{u}\:\right)+{C} \\ $$$$=\frac{\mathrm{1}}{\mathrm{12}}{sh}^{\mathrm{2}} {u}\:{shu}\:−\frac{\mathrm{1}}{\mathrm{16}}{sh}\left(\mathrm{2}{u}\right)+\frac{{u}}{\mathrm{8}}\:+{C} \\ $$$$=\frac{\mathrm{1}}{\mathrm{12}}\left(\left(\mathrm{2}{t}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{1}\right)\sqrt{\left(\mathrm{2}{t}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{8}}\left(\mathrm{2}{t}+\mathrm{1}\right)\sqrt{\left(\mathrm{2}{t}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{8}}{argch}\left(\mathrm{2}{t}+\mathrm{1}\right)\:+{C} \\ $$$$=\frac{\mathrm{1}}{\mathrm{12}}\left(\mathrm{4}{t}^{\mathrm{2}} +\mathrm{4}{t}\right)\sqrt{\mathrm{4}{t}^{\mathrm{2}} +\mathrm{4}{t}}−\frac{\mathrm{1}}{\mathrm{8}}\left(\mathrm{2}{t}+\mathrm{1}\right)\sqrt{\mathrm{4}{t}^{\mathrm{2}} +\mathrm{4}{t}}+\frac{\mathrm{1}}{\mathrm{8}}{ln}\left(\mathrm{2}{t}+\mathrm{1}+\sqrt{\left(\mathrm{2}{t}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{1}}\right)\:+{C} \\ $$$$\left.=\frac{\mathrm{2}}{\mathrm{3}}\left({t}^{\mathrm{2}} \:+{t}\right)\sqrt{{t}^{\mathrm{2}} +{t}}−\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{2}{t}+\mathrm{1}\right)\sqrt{{t}^{\mathrm{2}} \:+{t}}\:+\frac{\mathrm{1}}{\mathrm{8}}{ln}\left(\mathrm{2}{t}+\mathrm{1}\right)+\mathrm{2}\sqrt{{t}^{\mathrm{2}} \:+{t}}\right)+{C} \\ $$$$\left.\Rightarrow{I}\:=\frac{\mathrm{2}}{\mathrm{3}}\left({x}+\sqrt{{x}}\right)\sqrt{{x}+\sqrt{{x}}}−\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{2}\sqrt{{x}}+\mathrm{1}\right)\sqrt{{x}+\sqrt{{x}}}+\frac{\mathrm{1}}{\mathrm{8}}{ln}\left(\mathrm{2}\sqrt{{x}}+\mathrm{1}+\mathrm{2}\sqrt{{x}+\sqrt{{x}}}\right)\:\right)\:+{C} \\ $$
Answered by john santu last updated on 22/Feb/20

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