Question Number 37893 by abdo mathsup 649 cc last updated on 19/Jun/18
$${calculate}\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{{x}+\sqrt{{x}+\mathrm{1}}}\:{dx}\:. \\ $$
Commented by math khazana by abdo last updated on 21/Jun/18
![let (√(x+(√(x+1)))) =t ⇒x+(√(x+1))=t^2 ⇒ (√(x+1))=t^2 −x ⇒x+1=(t^2 −x)^2 ⇒ x=t^4 −2xt^2 +x^2 −1 ⇒x^2 −(2t^2 +1)x +t^4 −1=0⇒ Δ =(2t^2 +1)^2 −4(t^4 −1) =4t^4 +4t^2 +1−4t^4 +4 =4t^2 +5 ⇒ x_1 =((2t^2 +1+(√(4t^2 +5)))/2) x_2 =((2t^2 +1−(√(4t^2 +5)))/2) with condition t≥0 andx≥−1 and t^2 −x≥0 t^2 −x_2 =t^2 −((2t^2 +1−(√(4t^2 +5)))/2) =((−1 +(√(4t^2 +5)))/2) ≥0 x_2 +1= ((2t^2 +1−(√(4t^2 +5)))/2) +1 =((2t^2 +3 −(√(4t^2 +5)))/2) ≥0 so we take x=((2t^2 +1−(√(4t^2 +5)))/2) ⇒(dx/dt)=2t −(1/2) ((8t)/(2(√(4t^2 +5)))) =2t −((2t)/( (√(4t^2 +5)))) ⇒ ∫_0 ^1 (√(x+(√(x+1))))dx= ∫_1 ^(√(1+(√2))) t( 2t−((2t)/( (√(4t^2 +5)))))dt =2 ∫_1 ^(√(1+(√2))) t^2 dt −2 ∫_1 ^(√(1+(√2))) (t^2 /( (√(4t^2 +5))))dt =2[(t^3 /3)]_1 ^(√(1+(√2))) −(1/2) ∫_1 ^(√(1+(√2))) ((4t^2 +5 −5)/( (√(4t^2 +5))))dt =2( (1+(√2))(√(1+(√2))) −(1/3)) −(1/2) ∫_1 ^(√(1+(√2))) (√(4t^2 +5)) dt +(5/2) ∫_1 ^(√(1+(√2))) (dt/( (√(4t^2 +5)))) changement 2t =(√5) sh(x) give ∫_1 ^(√(1+(√2))) (√(4t^2 +5))dt = ∫_(argsh((2/( (√5))))) ^(arsh((2/( (√5)))(√(1+(√2))))) (√5)ch(x)((√5)/2)chx dx = (5/2)∫_α ^β ((1+ch(2x))/2)dx=(5/4)(β−α) +(5/8)[sh(2x)]_α ^β =(5/4)(β−α) +(5/8)(sh(2β)−sh(2α))....](https://www.tinkutara.com/question/Q38073.png)
$${let}\:\sqrt{{x}+\sqrt{{x}+\mathrm{1}}}\:\:={t}\:\Rightarrow{x}+\sqrt{{x}+\mathrm{1}}={t}^{\mathrm{2}} \:\Rightarrow \\ $$$$\sqrt{{x}+\mathrm{1}}={t}^{\mathrm{2}} \:−{x}\:\Rightarrow{x}+\mathrm{1}=\left({t}^{\mathrm{2}} −{x}\right)^{\mathrm{2}} \:\Rightarrow \\ $$$${x}={t}^{\mathrm{4}} \:−\mathrm{2}{xt}^{\mathrm{2}} \:+{x}^{\mathrm{2}} −\mathrm{1}\:\Rightarrow{x}^{\mathrm{2}} \:−\left(\mathrm{2}{t}^{\mathrm{2}} +\mathrm{1}\right){x}\:+{t}^{\mathrm{4}} −\mathrm{1}=\mathrm{0}\Rightarrow \\ $$$$\Delta\:=\left(\mathrm{2}{t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} \:−\mathrm{4}\left({t}^{\mathrm{4}} −\mathrm{1}\right) \\ $$$$=\mathrm{4}{t}^{\mathrm{4}} \:+\mathrm{4}{t}^{\mathrm{2}} \:+\mathrm{1}−\mathrm{4}{t}^{\mathrm{4}} \:+\mathrm{4}\:=\mathrm{4}{t}^{\mathrm{2}} \:+\mathrm{5}\:\Rightarrow \\ $$$${x}_{\mathrm{1}} =\frac{\mathrm{2}{t}^{\mathrm{2}} \:+\mathrm{1}+\sqrt{\mathrm{4}{t}^{\mathrm{2}} \:+\mathrm{5}}}{\mathrm{2}} \\ $$$${x}_{\mathrm{2}} =\frac{\mathrm{2}{t}^{\mathrm{2}} \:+\mathrm{1}−\sqrt{\mathrm{4}{t}^{\mathrm{2}} +\mathrm{5}}}{\mathrm{2}}\:\:{with}\:{condition}\:{t}\geqslant\mathrm{0}\:{andx}\geqslant−\mathrm{1} \\ $$$${and}\:{t}^{\mathrm{2}} −{x}\geqslant\mathrm{0} \\ $$$${t}^{\mathrm{2}} −{x}_{\mathrm{2}} ={t}^{\mathrm{2}} \:−\frac{\mathrm{2}{t}^{\mathrm{2}} \:+\mathrm{1}−\sqrt{\mathrm{4}{t}^{\mathrm{2}} \:+\mathrm{5}}}{\mathrm{2}} \\ $$$$=\frac{−\mathrm{1}\:+\sqrt{\mathrm{4}{t}^{\mathrm{2}} +\mathrm{5}}}{\mathrm{2}}\:\geqslant\mathrm{0} \\ $$$${x}_{\mathrm{2}} +\mathrm{1}=\:\frac{\mathrm{2}{t}^{\mathrm{2}} +\mathrm{1}−\sqrt{\mathrm{4}{t}^{\mathrm{2}} \:+\mathrm{5}}}{\mathrm{2}}\:+\mathrm{1} \\ $$$$=\frac{\mathrm{2}{t}^{\mathrm{2}} \:+\mathrm{3}\:−\sqrt{\mathrm{4}{t}^{\mathrm{2}} \:+\mathrm{5}}}{\mathrm{2}}\:\geqslant\mathrm{0}\:{so}\:{we}\:{take}\: \\ $$$${x}=\frac{\mathrm{2}{t}^{\mathrm{2}} \:+\mathrm{1}−\sqrt{\mathrm{4}{t}^{\mathrm{2}} \:+\mathrm{5}}}{\mathrm{2}}\:\Rightarrow\frac{{dx}}{{dt}}=\mathrm{2}{t}\:−\frac{\mathrm{1}}{\mathrm{2}}\:\:\frac{\mathrm{8}{t}}{\mathrm{2}\sqrt{\mathrm{4}{t}^{\mathrm{2}} +\mathrm{5}}} \\ $$$$=\mathrm{2}{t}\:\:\:−\frac{\mathrm{2}{t}}{\:\sqrt{\mathrm{4}{t}^{\mathrm{2}} +\mathrm{5}}}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{{x}+\sqrt{{x}+\mathrm{1}}}{dx}=\:\int_{\mathrm{1}} ^{\sqrt{\mathrm{1}+\sqrt{\mathrm{2}}}} \:\:{t}\left(\:\mathrm{2}{t}−\frac{\mathrm{2}{t}}{\:\sqrt{\mathrm{4}{t}^{\mathrm{2}} +\mathrm{5}}}\right){dt} \\ $$$$=\mathrm{2}\:\int_{\mathrm{1}} ^{\sqrt{\mathrm{1}+\sqrt{\mathrm{2}}}} {t}^{\mathrm{2}} {dt}\:\:−\mathrm{2}\:\int_{\mathrm{1}} ^{\sqrt{\mathrm{1}+\sqrt{\mathrm{2}}}} \:\:\frac{{t}^{\mathrm{2}} }{\:\sqrt{\mathrm{4}{t}^{\mathrm{2}} \:+\mathrm{5}}}{dt} \\ $$$$=\mathrm{2}\left[\frac{{t}^{\mathrm{3}} }{\mathrm{3}}\right]_{\mathrm{1}} ^{\sqrt{\mathrm{1}+\sqrt{\mathrm{2}}}} \:\:−\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{1}} ^{\sqrt{\mathrm{1}+\sqrt{\mathrm{2}}}} \:\:\frac{\mathrm{4}{t}^{\mathrm{2}} \:+\mathrm{5}\:−\mathrm{5}}{\:\sqrt{\mathrm{4}{t}^{\mathrm{2}} +\mathrm{5}}}{dt} \\ $$$$=\mathrm{2}\left(\:\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)\sqrt{\mathrm{1}+\sqrt{\mathrm{2}}}\:−\frac{\mathrm{1}}{\mathrm{3}}\right)\:−\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{1}} ^{\sqrt{\mathrm{1}+\sqrt{\mathrm{2}}}} \sqrt{\mathrm{4}{t}^{\mathrm{2}} \:+\mathrm{5}}\:{dt} \\ $$$$+\frac{\mathrm{5}}{\mathrm{2}}\:\int_{\mathrm{1}} ^{\sqrt{\mathrm{1}+\sqrt{\mathrm{2}}}} \:\:\:\frac{{dt}}{\:\sqrt{\mathrm{4}{t}^{\mathrm{2}} \:+\mathrm{5}}}\:\:{changement} \\ $$$$\mathrm{2}{t}\:=\sqrt{\mathrm{5}}\:{sh}\left({x}\right)\:{give} \\ $$$$\int_{\mathrm{1}} ^{\sqrt{\mathrm{1}+\sqrt{\mathrm{2}}}} \:\:\sqrt{\mathrm{4}{t}^{\mathrm{2}} \:+\mathrm{5}}{dt}\:=\:\int_{{argsh}\left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}}\right)} ^{{arsh}\left(\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}}\sqrt{\mathrm{1}+\sqrt{\mathrm{2}}}\right)} \:\:\:\sqrt{\mathrm{5}}{ch}\left({x}\right)\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}{chx}\:{dx} \\ $$$$=\:\frac{\mathrm{5}}{\mathrm{2}}\int_{\alpha} ^{\beta} \:\:\:\:\:\frac{\mathrm{1}+{ch}\left(\mathrm{2}{x}\right)}{\mathrm{2}}{dx}=\frac{\mathrm{5}}{\mathrm{4}}\left(\beta−\alpha\right)\:+\frac{\mathrm{5}}{\mathrm{8}}\left[{sh}\left(\mathrm{2}{x}\right)\right]_{\alpha} ^{\beta} \\ $$$$=\frac{\mathrm{5}}{\mathrm{4}}\left(\beta−\alpha\right)\:+\frac{\mathrm{5}}{\mathrm{8}}\left({sh}\left(\mathrm{2}\beta\right)−{sh}\left(\mathrm{2}\alpha\right)\right)…. \\ $$
Answered by MJS last updated on 19/Jun/18
![∫(√(x+(√(x+1))))dx= [t=(√(x+1)) → dx=2(√(x+1))dt] =2∫t(√(t^2 +t−1))dt= =∫(2t+1)(√(t^2 +t−1))dt−∫(√(t^2 +t−1))dt= ∫(2t+1)(√(t^2 +t−1))dt= [u=t^2 +t−1 → dt=(du/(2t+1))] =∫(√u)du=(2/3)(√u^3 )=(2/3)(√((t^2 +t−1)^3 ))= =(2/3)(√((x+(√(x+1)))^3 )) ∫(√(t^2 +t−1))dt=∫(√((t+(1/2))^2 −(5/4)))dt= =(1/2)∫(√((2t+1)^2 −5))dt= [v=2t+1 → dt=(dv/2)] =(1/4)∫(√(v^2 −5))dv= =(1/4)((1/2)(v(√(v^2 −5))−5ln(v+(√(v^2 −5)))))= =(1/8)(2t+1)(√((2t+1)^2 −5))− −(5/8)ln(2t+1+(√((2t+1)^2 −5)))= =(1/4)(1+2(√(x+1)))(√(x+(√(x+1))))− −(5/8)ln(1+2(√(x+1))+2(√(x+(√(x+1))))) =(1/(12))(√(x+(√(x+1))))(8x−3+2(√(x+1)))+(5/8)ln(1+2(√(x+1))+2(√(x+(√(x+1)))))+C](https://www.tinkutara.com/question/Q37917.png)
$$\int\sqrt{{x}+\sqrt{{x}+\mathrm{1}}}{dx}= \\ $$$$\:\:\:\:\:\:\:\:\:\:\left[{t}=\sqrt{{x}+\mathrm{1}}\:\rightarrow\:{dx}=\mathrm{2}\sqrt{{x}+\mathrm{1}}{dt}\right] \\ $$$$=\mathrm{2}\int{t}\sqrt{{t}^{\mathrm{2}} +{t}−\mathrm{1}}{dt}= \\ $$$$=\int\left(\mathrm{2}{t}+\mathrm{1}\right)\sqrt{{t}^{\mathrm{2}} +{t}−\mathrm{1}}{dt}−\int\sqrt{{t}^{\mathrm{2}} +{t}−\mathrm{1}}{dt}= \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\int\left(\mathrm{2}{t}+\mathrm{1}\right)\sqrt{{t}^{\mathrm{2}} +{t}−\mathrm{1}}{dt}= \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left[{u}={t}^{\mathrm{2}} +{t}−\mathrm{1}\:\rightarrow\:{dt}=\frac{{du}}{\mathrm{2}{t}+\mathrm{1}}\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:=\int\sqrt{{u}}{du}=\frac{\mathrm{2}}{\mathrm{3}}\sqrt{{u}^{\mathrm{3}} }=\frac{\mathrm{2}}{\mathrm{3}}\sqrt{\left({t}^{\mathrm{2}} +{t}−\mathrm{1}\right)^{\mathrm{3}} }= \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{2}}{\mathrm{3}}\sqrt{\left({x}+\sqrt{{x}+\mathrm{1}}\right)^{\mathrm{3}} } \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\int\sqrt{{t}^{\mathrm{2}} +{t}−\mathrm{1}}{dt}=\int\sqrt{\left({t}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} −\frac{\mathrm{5}}{\mathrm{4}}}{dt}= \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\int\sqrt{\left(\mathrm{2}{t}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{5}}{dt}= \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left[{v}=\mathrm{2}{t}+\mathrm{1}\:\rightarrow\:{dt}=\frac{{dv}}{\mathrm{2}}\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{4}}\int\sqrt{{v}^{\mathrm{2}} −\mathrm{5}}{dv}= \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\mathrm{1}}{\mathrm{2}}\left({v}\sqrt{{v}^{\mathrm{2}} −\mathrm{5}}−\mathrm{5ln}\left({v}+\sqrt{{v}^{\mathrm{2}} −\mathrm{5}}\right)\right)\right)= \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{8}}\left(\mathrm{2}{t}+\mathrm{1}\right)\sqrt{\left(\mathrm{2}{t}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{5}}− \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\frac{\mathrm{5}}{\mathrm{8}}\mathrm{ln}\left(\mathrm{2}{t}+\mathrm{1}+\sqrt{\left(\mathrm{2}{t}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{5}}\right)= \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{1}+\mathrm{2}\sqrt{{x}+\mathrm{1}}\right)\sqrt{{x}+\sqrt{{x}+\mathrm{1}}}− \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\frac{\mathrm{5}}{\mathrm{8}}\mathrm{ln}\left(\mathrm{1}+\mathrm{2}\sqrt{{x}+\mathrm{1}}+\mathrm{2}\sqrt{{x}+\sqrt{{x}+\mathrm{1}}}\right) \\ $$$$ \\ $$$$=\frac{\mathrm{1}}{\mathrm{12}}\sqrt{{x}+\sqrt{{x}+\mathrm{1}}}\left(\mathrm{8}{x}−\mathrm{3}+\mathrm{2}\sqrt{{x}+\mathrm{1}}\right)+\frac{\mathrm{5}}{\mathrm{8}}\mathrm{ln}\left(\mathrm{1}+\mathrm{2}\sqrt{{x}+\mathrm{1}}+\mathrm{2}\sqrt{{x}+\sqrt{{x}+\mathrm{1}}}\right)+{C} \\ $$
Commented by prof Abdo imad last updated on 19/Jun/18
$${thanks}\:\:{sir}\:{Mjs}\:{you}\:{are}\:{really}\:{talented}. \\ $$