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Question Number 133674 by liberty last updated on 23/Feb/21
∫ (√(1+(√(1+(√x))))) dx =?
$$\int\:\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+\sqrt{\mathrm{x}}}}\:\mathrm{dx}\:=?\: \\ $$
Answered by EDWIN88 last updated on 23/Feb/21
let y = (√(1+(√(1+(√x))))) ⇒(√(1+(√x))) = y^2 −1 ⇒1+(√x) = y^4 −2y^2 +1 , x = (y^4 −2y^2 )^2 dx = 2(4y^3 −4y)(y^4 −2y^2 ) I=∫ y(4y^3 −8y)(y^4 −2y^2 )dy I=∫(4y^4 −8y^2 )(y^4 −2y^2 )dy I=∫ (4y^8 −16y^6 +16y^4 )dy I=(4/9)y^9 −((16)/7)y^7 +((16)/5)y^5 +c I=(4/9)((√(1+(√(1+(√x))))))^9 −((16)/7)((√(1+(√(1+(√x))))))^7 +((16)/5)((√(1+(√(1+(√x))))))^5 + c
$$\:\mathrm{let}\:\mathrm{y}\:=\:\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+\sqrt{\mathrm{x}}}}\:\Rightarrow\sqrt{\mathrm{1}+\sqrt{\mathrm{x}}}\:=\:\mathrm{y}^{\mathrm{2}} −\mathrm{1} \\ $$$$\Rightarrow\mathrm{1}+\sqrt{\mathrm{x}}\:=\:\mathrm{y}^{\mathrm{4}} −\mathrm{2y}^{\mathrm{2}} +\mathrm{1}\:,\:\mathrm{x}\:=\:\left(\mathrm{y}^{\mathrm{4}} −\mathrm{2y}^{\mathrm{2}} \right)^{\mathrm{2}} \\ $$$$\:\mathrm{dx}\:=\:\mathrm{2}\left(\mathrm{4y}^{\mathrm{3}} −\mathrm{4y}\right)\left(\mathrm{y}^{\mathrm{4}} −\mathrm{2y}^{\mathrm{2}} \right) \\ $$$$ \\ $$$$\mathrm{I}=\int\:\mathrm{y}\left(\mathrm{4y}^{\mathrm{3}} −\mathrm{8y}\right)\left(\mathrm{y}^{\mathrm{4}} −\mathrm{2y}^{\mathrm{2}} \right)\mathrm{dy} \\ $$$$\mathrm{I}=\int\left(\mathrm{4y}^{\mathrm{4}} −\mathrm{8y}^{\mathrm{2}} \right)\left(\mathrm{y}^{\mathrm{4}} −\mathrm{2y}^{\mathrm{2}} \right)\mathrm{dy} \\ $$$$\mathrm{I}=\int\:\left(\mathrm{4y}^{\mathrm{8}} −\mathrm{16y}^{\mathrm{6}} +\mathrm{16y}^{\mathrm{4}} \right)\mathrm{dy}\: \\ $$$$\mathrm{I}=\frac{\mathrm{4}}{\mathrm{9}}\mathrm{y}^{\mathrm{9}} −\frac{\mathrm{16}}{\mathrm{7}}\mathrm{y}^{\mathrm{7}} +\frac{\mathrm{16}}{\mathrm{5}}\mathrm{y}^{\mathrm{5}} +\mathrm{c} \\ $$$$\mathrm{I}=\frac{\mathrm{4}}{\mathrm{9}}\left(\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+\sqrt{\mathrm{x}}}}\right)^{\mathrm{9}} −\frac{\mathrm{16}}{\mathrm{7}}\left(\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+\sqrt{\mathrm{x}}}}\right)^{\mathrm{7}} +\frac{\mathrm{16}}{\mathrm{5}}\left(\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+\sqrt{\mathrm{x}}}}\right)^{\mathrm{5}} \:+\:\mathrm{c} \\ $$
Answered by Olaf last updated on 23/Feb/21
F(x) = ∫(√(1+(√(1+(√(x ))))))dx Let u = 1+(√x) du = (dx/(2(√x))) = (dx/(2(u−1))) F(x) = ∫(√(1+(√u))) 2(u−1)du Let v = 1+(√u) dv = (du/(2(√x))) = (du/(2(v−1))) F(x) = ∫(√v) 2[(v−1)^2 −1]2(v−1)dv F(x) = 4∫(√v) [(v−1)^3 −v+1]dv F(x) = 4∫(√v) [v^3 −3v^2 +2v]dv F(x) = 4∫(v^(7/2) −3v^(5/2) +2v^(3/2) )dv F(x) = 4((2/9)v^(9/2) −(6/7)v^(7/2) +(4/5)v^(5/2) )+C F(x) = 4v^(5/2) ((2/9)v^2 −(6/7)v+(4/5))+C F(x) = 4[1+(√(1+(√x)))]^(5/2) [(2/9)(1+(√(1+(√x))))^2 −(6/7)(1+(√(1+(√x))))+(4/5)]+C F(x) = 4[1+(√(1+(√x)))]^(5/2) [(2/9)(√x)−((26)/(63))(√(1+(√x)))−((122)/(315))]+C
$$\mathrm{F}\left({x}\right)\:=\:\int\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+\sqrt{{x}\:}}}{dx} \\ $$$$\mathrm{Let}\:{u}\:=\:\mathrm{1}+\sqrt{{x}} \\ $$$${du}\:=\:\frac{{dx}}{\mathrm{2}\sqrt{{x}}}\:=\:\frac{{dx}}{\mathrm{2}\left({u}−\mathrm{1}\right)} \\ $$$$\mathrm{F}\left({x}\right)\:=\:\int\sqrt{\mathrm{1}+\sqrt{{u}}}\:\mathrm{2}\left({u}−\mathrm{1}\right){du} \\ $$$$\mathrm{Let}\:{v}\:=\:\mathrm{1}+\sqrt{{u}} \\ $$$${dv}\:=\:\frac{{du}}{\mathrm{2}\sqrt{{x}}}\:=\:\frac{{du}}{\mathrm{2}\left({v}−\mathrm{1}\right)} \\ $$$$\mathrm{F}\left({x}\right)\:=\:\int\sqrt{{v}}\:\mathrm{2}\left[\left({v}−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{1}\right]\mathrm{2}\left({v}−\mathrm{1}\right){dv} \\ $$$$\mathrm{F}\left({x}\right)\:=\:\mathrm{4}\int\sqrt{{v}}\:\left[\left({v}−\mathrm{1}\right)^{\mathrm{3}} −{v}+\mathrm{1}\right]{dv} \\ $$$$\mathrm{F}\left({x}\right)\:=\:\mathrm{4}\int\sqrt{{v}}\:\left[{v}^{\mathrm{3}} −\mathrm{3}{v}^{\mathrm{2}} +\mathrm{2}{v}\right]{dv} \\ $$$$\mathrm{F}\left({x}\right)\:=\:\mathrm{4}\int\left({v}^{\frac{\mathrm{7}}{\mathrm{2}}} −\mathrm{3}{v}^{\frac{\mathrm{5}}{\mathrm{2}}} +\mathrm{2}{v}^{\frac{\mathrm{3}}{\mathrm{2}}} \right){dv} \\ $$$$\mathrm{F}\left({x}\right)\:=\:\mathrm{4}\left(\frac{\mathrm{2}}{\mathrm{9}}{v}^{\frac{\mathrm{9}}{\mathrm{2}}} −\frac{\mathrm{6}}{\mathrm{7}}{v}^{\frac{\mathrm{7}}{\mathrm{2}}} +\frac{\mathrm{4}}{\mathrm{5}}{v}^{\frac{\mathrm{5}}{\mathrm{2}}} \right)+\mathrm{C} \\ $$$$\mathrm{F}\left({x}\right)\:=\:\mathrm{4}{v}^{\frac{\mathrm{5}}{\mathrm{2}}} \left(\frac{\mathrm{2}}{\mathrm{9}}{v}^{\mathrm{2}} −\frac{\mathrm{6}}{\mathrm{7}}{v}+\frac{\mathrm{4}}{\mathrm{5}}\right)+\mathrm{C} \\ $$$$\mathrm{F}\left({x}\right)\:=\:\mathrm{4}\left[\mathrm{1}+\sqrt{\mathrm{1}+\sqrt{{x}}}\right]^{\frac{\mathrm{5}}{\mathrm{2}}} \left[\frac{\mathrm{2}}{\mathrm{9}}\left(\mathrm{1}+\sqrt{\mathrm{1}+\sqrt{{x}}}\right)^{\mathrm{2}} −\frac{\mathrm{6}}{\mathrm{7}}\left(\mathrm{1}+\sqrt{\mathrm{1}+\sqrt{{x}}}\right)+\frac{\mathrm{4}}{\mathrm{5}}\right]+\mathrm{C} \\ $$$$\mathrm{F}\left({x}\right)\:=\:\mathrm{4}\left[\mathrm{1}+\sqrt{\mathrm{1}+\sqrt{{x}}}\right]^{\frac{\mathrm{5}}{\mathrm{2}}} \left[\frac{\mathrm{2}}{\mathrm{9}}\sqrt{{x}}−\frac{\mathrm{26}}{\mathrm{63}}\sqrt{\mathrm{1}+\sqrt{{x}}}−\frac{\mathrm{122}}{\mathrm{315}}\right]+\mathrm{C} \\ $$

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