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Question Number 56700 by maxmathsup by imad last updated on 21/Mar/19
find ∫ (√(x−2(√x)+3))dx
$$\:{find}\:\int\:\sqrt{{x}−\mathrm{2}\sqrt{{x}}+\mathrm{3}}{dx} \\ $$
Commented by kaivan.ahmadi last updated on 22/Mar/19
Commented by maxmathsup by imad last updated on 23/Mar/19
I =∫(√(((√x)−1)^2 +2))dx changement (√x)−1=(√2)sh(t) give (√x)=1+(√2)sht ⇒ x =(1+(√2)sh(t))^2 ⇒ I =∫ (√(2sh^2 t +2))(2(1+(√2)sh(t))(√2)cht dt =4 ∫ ch^2 (t)(1+(√2)sh(t)) dt =4 ∫ ch^2 t dt +4(√2)∫ sh(t)ch^2 t dt =4 ∫ ((1+ch(2t))/2)dt +4(√2)(1/3)ch^3 (t) =2t +2 ∫ ch(2t)dt +((4(√2))/3) ch^3 (t) =2t +sh(2t) +((4(√2))/3) ch^3 (t) but t =argsh((((√x)−1)/2)) =ln((((√x)−1)/2) +(√(1+((((√x)−1)/2))^2 ))) I =2argsh((((√x)−1)/2)) +sh(2 argsh((((√x)−1)/2))) +((4(√2))/3) ch^3 (argsh((((√x)−1)/2))) +c
$${I}\:=\int\sqrt{\left(\sqrt{{x}}−\mathrm{1}\right)^{\mathrm{2}} \:+\mathrm{2}}{dx}\:\:\:{changement}\:\sqrt{{x}}−\mathrm{1}=\sqrt{\mathrm{2}}{sh}\left({t}\right)\:{give}\:\sqrt{{x}}=\mathrm{1}+\sqrt{\mathrm{2}}{sht}\:\Rightarrow \\ $$$${x}\:=\left(\mathrm{1}+\sqrt{\mathrm{2}}{sh}\left({t}\right)\right)^{\mathrm{2}} \:\Rightarrow \\ $$$${I}\:=\int\:\sqrt{\mathrm{2}{sh}^{\mathrm{2}} {t}\:+\mathrm{2}}\left(\mathrm{2}\left(\mathrm{1}+\sqrt{\mathrm{2}}{sh}\left({t}\right)\right)\sqrt{\mathrm{2}}{cht}\:{dt}\right. \\ $$$$=\mathrm{4}\:\int\:{ch}^{\mathrm{2}} \left({t}\right)\left(\mathrm{1}+\sqrt{\mathrm{2}}{sh}\left({t}\right)\right)\:{dt}\:=\mathrm{4}\:\int\:{ch}^{\mathrm{2}} {t}\:{dt}\:\:+\mathrm{4}\sqrt{\mathrm{2}}\int\:\:{sh}\left({t}\right){ch}^{\mathrm{2}} {t}\:{dt} \\ $$$$=\mathrm{4}\:\int\:\frac{\mathrm{1}+{ch}\left(\mathrm{2}{t}\right)}{\mathrm{2}}{dt}\:\:+\mathrm{4}\sqrt{\mathrm{2}}\frac{\mathrm{1}}{\mathrm{3}}{ch}^{\mathrm{3}} \left({t}\right)\:=\mathrm{2}{t}\:+\mathrm{2}\:\int\:{ch}\left(\mathrm{2}{t}\right){dt}\:+\frac{\mathrm{4}\sqrt{\mathrm{2}}}{\mathrm{3}}\:{ch}^{\mathrm{3}} \left({t}\right) \\ $$$$=\mathrm{2}{t}\:+{sh}\left(\mathrm{2}{t}\right)\:+\frac{\mathrm{4}\sqrt{\mathrm{2}}}{\mathrm{3}}\:{ch}^{\mathrm{3}} \left({t}\right)\:\:{but}\:{t}\:={argsh}\left(\frac{\sqrt{{x}}−\mathrm{1}}{\mathrm{2}}\right)\:={ln}\left(\frac{\sqrt{{x}}−\mathrm{1}}{\mathrm{2}}\:+\sqrt{\mathrm{1}+\left(\frac{\sqrt{{x}}−\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }\right) \\ $$$${I}\:=\mathrm{2}{argsh}\left(\frac{\sqrt{{x}}−\mathrm{1}}{\mathrm{2}}\right)\:+{sh}\left(\mathrm{2}\:{argsh}\left(\frac{\sqrt{{x}}−\mathrm{1}}{\mathrm{2}}\right)\right)\:+\frac{\mathrm{4}\sqrt{\mathrm{2}}}{\mathrm{3}}\:{ch}^{\mathrm{3}} \left({argsh}\left(\frac{\sqrt{{x}}−\mathrm{1}}{\mathrm{2}}\right)\right)\:+{c} \\ $$
Answered by kaivan.ahmadi last updated on 22/Mar/19
∫(√(((√x)−1)^2 +2))dx= t=(√x)−1⇒dt=(dx/(2(√x)))=(dx/(2t+2))⇒dx=(2t+2)dt ∫(√(t^2 +2))(2t+2)dt=∫2t(√(t^2 +2))dt+∫2(√(t^2 +2))dt=I+J solve I u=t^2 +2⇒du=2tdt ∫(√u)du=(2/3)u^(3/2) +C_1 =(2/3)(t^2 +2)^(3/2) +C_1 = (((√x)−1)^2 +2)^(3/2) +C_1 solve J t=(√2)tgθ⇒dt=(√2)(1+tg^2 θ)dθ and so tgθ=(t/( (√2)))⇒secθ=(√(1+(t^2 /2))) ∫2(√(2(1+tg^2 θ))).(√2)(1+tg^2 θ)dθ= ∫4(1+tg^2 θ)(√(1+tg^2 θ))dθ=∫4(dθ/(cos^3 θ))= 4∫sec^3 θdθ=4((1/2)secθtgθ+(1/2)(secθ+tgθ))+C_2 = 2secθtgθ+2(secθ+tgθ)+C_2 = 2(√(1+(t^2 /2))).(t/( (√2)))+2((√(1+(t^2 /2)))+(t/( (√2))))+C_2
$$\int\sqrt{\left(\sqrt{{x}}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{2}}{dx}= \\ $$$${t}=\sqrt{{x}}−\mathrm{1}\Rightarrow{dt}=\frac{{dx}}{\mathrm{2}\sqrt{{x}}}=\frac{{dx}}{\mathrm{2}{t}+\mathrm{2}}\Rightarrow{dx}=\left(\mathrm{2}{t}+\mathrm{2}\right){dt} \\ $$$$\int\sqrt{{t}^{\mathrm{2}} +\mathrm{2}}\left(\mathrm{2}{t}+\mathrm{2}\right){dt}=\int\mathrm{2}{t}\sqrt{{t}^{\mathrm{2}} +\mathrm{2}}{dt}+\int\mathrm{2}\sqrt{{t}^{\mathrm{2}} +\mathrm{2}}{dt}={I}+{J} \\ $$$${solve}\:{I} \\ $$$${u}={t}^{\mathrm{2}} +\mathrm{2}\Rightarrow{du}=\mathrm{2}{tdt} \\ $$$$\int\sqrt{{u}}{du}=\frac{\mathrm{2}}{\mathrm{3}}{u}^{\frac{\mathrm{3}}{\mathrm{2}}} +{C}_{\mathrm{1}} =\frac{\mathrm{2}}{\mathrm{3}}\left({t}^{\mathrm{2}} +\mathrm{2}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} +{C}_{\mathrm{1}} = \\ $$$$\left(\left(\sqrt{{x}}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{2}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} +{C}_{\mathrm{1}} \\ $$$${solve}\:{J} \\ $$$${t}=\sqrt{\mathrm{2}}{tg}\theta\Rightarrow{dt}=\sqrt{\mathrm{2}}\left(\mathrm{1}+{tg}^{\mathrm{2}} \theta\right){d}\theta \\ $$$${and}\:{so}\:{tg}\theta=\frac{{t}}{\:\sqrt{\mathrm{2}}}\Rightarrow{sec}\theta=\sqrt{\mathrm{1}+\frac{{t}^{\mathrm{2}} }{\mathrm{2}}} \\ $$$$\int\mathrm{2}\sqrt{\mathrm{2}\left(\mathrm{1}+{tg}^{\mathrm{2}} \theta\right)}.\sqrt{\mathrm{2}}\left(\mathrm{1}+{tg}^{\mathrm{2}} \theta\right){d}\theta= \\ $$$$\int\mathrm{4}\left(\mathrm{1}+{tg}^{\mathrm{2}} \theta\right)\sqrt{\mathrm{1}+{tg}^{\mathrm{2}} \theta}{d}\theta=\int\mathrm{4}\frac{{d}\theta}{{cos}^{\mathrm{3}} \theta}= \\ $$$$\mathrm{4}\int{sec}^{\mathrm{3}} \theta{d}\theta=\mathrm{4}\left(\frac{\mathrm{1}}{\mathrm{2}}{sec}\theta{tg}\theta+\frac{\mathrm{1}}{\mathrm{2}}\left({sec}\theta+{tg}\theta\right)\right)+{C}_{\mathrm{2}} = \\ $$$$\mathrm{2}{sec}\theta{tg}\theta+\mathrm{2}\left({sec}\theta+{tg}\theta\right)+{C}_{\mathrm{2}} = \\ $$$$\mathrm{2}\sqrt{\mathrm{1}+\frac{{t}^{\mathrm{2}} }{\mathrm{2}}}.\frac{{t}}{\:\sqrt{\mathrm{2}}}+\mathrm{2}\left(\sqrt{\mathrm{1}+\frac{{t}^{\mathrm{2}} }{\mathrm{2}}}+\frac{{t}}{\:\sqrt{\mathrm{2}}}\right)+{C}_{\mathrm{2}} \\ $$$$ \\ $$

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