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Question Number 95371 by Fikret last updated on 24/May/20
∫(x^(2/3) /( (√(1+x^(2/3) ))))dx=?
$$\int\frac{{x}^{\mathrm{2}/\mathrm{3}} }{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}/\mathrm{3}} }}{dx}=? \\ $$
Commented by bobhans last updated on 25/May/20
∫ ((x^2 )^(1/(3 )) /( (√(1+(x^2 )^(1/(3 )) )))) dx . let (√(1+(x^2 )^(1/(3 )) )) = t 1+(x^2 )^(1/(3 )) = t^2 ⇒(x^2 )^(1/(3 )) = t^2 −1 ; (x)^(1/(3 )) = (√(t^2 −1)) ((2 dx)/(3 (x)^(1/(3 )) )) = 2t dt ⇒ dx = 3t(√(t^2 −1)) dt J = ∫ (( (t^2 −1)3t(√(t^2 −1)))/(t )) = ∫ 3(t^2 −1)^(3/2) dt let again t=sec θ J = 3∫tan^4 θ sec θ dθ
$$\int\:\frac{\sqrt[{\mathrm{3}\:\:}]{\mathrm{x}^{\mathrm{2}} }}{\:\sqrt{\mathrm{1}+\sqrt[{\mathrm{3}\:\:}]{\mathrm{x}^{\mathrm{2}} }}}\:\mathrm{dx}\:.\:\mathrm{let}\:\sqrt{\mathrm{1}+\sqrt[{\mathrm{3}\:\:}]{\mathrm{x}^{\mathrm{2}} }}\:=\:\mathrm{t} \\ $$$$\mathrm{1}+\sqrt[{\mathrm{3}\:\:}]{\mathrm{x}^{\mathrm{2}} }\:=\:\mathrm{t}^{\mathrm{2}} \:\Rightarrow\sqrt[{\mathrm{3}\:\:}]{\mathrm{x}^{\mathrm{2}} }\:=\:\mathrm{t}^{\mathrm{2}} −\mathrm{1}\:;\:\sqrt[{\mathrm{3}\:\:}]{\mathrm{x}}\:=\:\sqrt{\mathrm{t}^{\mathrm{2}} −\mathrm{1}} \\ $$$$\frac{\mathrm{2}\:\mathrm{dx}}{\mathrm{3}\:\sqrt[{\mathrm{3}\:\:}]{\mathrm{x}}}\:=\:\mathrm{2t}\:\mathrm{dt}\:\Rightarrow\:\mathrm{dx}\:=\:\mathrm{3t}\sqrt{\mathrm{t}^{\mathrm{2}} −\mathrm{1}}\:\mathrm{dt}\: \\ $$$$\mathrm{J}\:=\:\int\:\frac{\:\left(\mathrm{t}^{\mathrm{2}} −\mathrm{1}\right)\mathrm{3t}\sqrt{\mathrm{t}^{\mathrm{2}} −\mathrm{1}}}{\mathrm{t}\:}\:=\:\int\:\mathrm{3}\left(\mathrm{t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{3}/\mathrm{2}} \:\mathrm{dt}\: \\ $$$$\mathrm{let}\:\mathrm{again}\:\mathrm{t}=\mathrm{sec}\:\theta\: \\ $$$$\mathrm{J}\:=\:\mathrm{3}\int\mathrm{tan}\:^{\mathrm{4}} \theta\:\mathrm{sec}\:\theta\:\mathrm{d}\theta\: \\ $$
Answered by MJS last updated on 25/May/20
∫(x^(2/3) /( (√(x^(2/3) +1))))dx= [t=x^(1/3) +(√(x^(2/3) +1)) → dx=3((x^(2/3) (√(x^(2/3) +1)))/(x^(1/3) +(√(x^(2/3) +1))))dt] =(3/(16))∫(((t^2 −1)^4 )/t^5 )dt= =∫((3/(16))t^3 −(3/4)t+(9/(8t))−(3/(4t^3 ))+(3/(16t^5 )))dt= =((3(t^4 −1)(t^4 −8t^2 +1))/(64t^4 ))+(9/8)ln t = =(3/8)(2x−3x^(1/3) )(√(x^(2/3) +1))+(9/8)ln (x^(1/3) +(√(x^(2/3) +1))) +C
$$\int\frac{{x}^{\mathrm{2}/\mathrm{3}} }{\:\sqrt{{x}^{\mathrm{2}/\mathrm{3}} +\mathrm{1}}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}={x}^{\mathrm{1}/\mathrm{3}} +\sqrt{{x}^{\mathrm{2}/\mathrm{3}} +\mathrm{1}}\:\rightarrow\:{dx}=\mathrm{3}\frac{{x}^{\mathrm{2}/\mathrm{3}} \sqrt{{x}^{\mathrm{2}/\mathrm{3}} +\mathrm{1}}}{{x}^{\mathrm{1}/\mathrm{3}} +\sqrt{{x}^{\mathrm{2}/\mathrm{3}} +\mathrm{1}}}{dt}\right] \\ $$$$=\frac{\mathrm{3}}{\mathrm{16}}\int\frac{\left({t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{4}} }{{t}^{\mathrm{5}} }{dt}= \\ $$$$=\int\left(\frac{\mathrm{3}}{\mathrm{16}}{t}^{\mathrm{3}} −\frac{\mathrm{3}}{\mathrm{4}}{t}+\frac{\mathrm{9}}{\mathrm{8}{t}}−\frac{\mathrm{3}}{\mathrm{4}{t}^{\mathrm{3}} }+\frac{\mathrm{3}}{\mathrm{16}{t}^{\mathrm{5}} }\right){dt}= \\ $$$$=\frac{\mathrm{3}\left({t}^{\mathrm{4}} −\mathrm{1}\right)\left({t}^{\mathrm{4}} −\mathrm{8}{t}^{\mathrm{2}} +\mathrm{1}\right)}{\mathrm{64}{t}^{\mathrm{4}} }+\frac{\mathrm{9}}{\mathrm{8}}\mathrm{ln}\:{t}\:= \\ $$$$=\frac{\mathrm{3}}{\mathrm{8}}\left(\mathrm{2}{x}−\mathrm{3}{x}^{\mathrm{1}/\mathrm{3}} \right)\sqrt{{x}^{\mathrm{2}/\mathrm{3}} +\mathrm{1}}+\frac{\mathrm{9}}{\mathrm{8}}\mathrm{ln}\:\left({x}^{\mathrm{1}/\mathrm{3}} +\sqrt{{x}^{\mathrm{2}/\mathrm{3}} +\mathrm{1}}\right)\:+\mathrm{C} \\ $$
Commented by Fikret last updated on 24/May/20
i didnt understand this i m sorry
$${i}\:{didnt}\:{understand}\:{this}\:{i}\:{m}\:{sorry} \\ $$
Commented by Fikret last updated on 24/May/20
Commented by Fikret last updated on 24/May/20
i didnt understood this
$${i}\:{didnt}\:{understood}\:{this} \\ $$
Answered by mathmax by abdo last updated on 25/May/20
=∫ (x^(2/3) /( (√(1+x^(2/3) ))))dx changement x^(1/3) =sh(t) ⇒x =sh^3 t ⇒ (dx/dt) =3 sh^2 t cht ⇒I =∫ ((sh^2 t)/(ch(t)))×3sh^2 t cht dt =3 ∫ sh^4 t dt =3 ∫ (((ch(2t)−1)/2))^2 dt =(3/4) ∫ (ch^2 (2t)−2ch(2t)+1)dt =(3/4) ∫ ch^2 (2t)dt −(3/2) ∫ ch(2t)dt +(3/4)t =(3/8)∫ (1+ch(4t))dt −(3/4)sh(2t) +(3/4)t =(3/8)t +(3/(32))sh(4t) −(3/4)sh(2t)+(3/4)t +C =(9/8)t +(3/(16))sh(2t)ch(2t)−(3/2)sh(t)ch(t) +C =(9/8)t +(3/8)sh(t)ch(t)(2sh^2 t+1)−(3/2)sh(t)cht +C I=(9/8)ln(x^(1/3) +(√(1+x^(2/3) ))) +(3/8) (^3 (√x))(√(1+(^3 (√x))^2 ))(2(x)^(2/3) +1) −(3/2)(^3 (√x))(√(1+(^3 (√x))^2 )) +C
$$\:=\int\:\frac{\mathrm{x}^{\frac{\mathrm{2}}{\mathrm{3}}} }{\:\sqrt{\mathrm{1}+\mathrm{x}^{\frac{\mathrm{2}}{\mathrm{3}}} }}\mathrm{dx}\:\:\mathrm{changement}\:\mathrm{x}^{\frac{\mathrm{1}}{\mathrm{3}}} \:=\mathrm{sh}\left(\mathrm{t}\right)\:\Rightarrow\mathrm{x}\:=\mathrm{sh}^{\mathrm{3}} \mathrm{t}\:\Rightarrow \\ $$$$\frac{\mathrm{dx}}{\mathrm{dt}}\:=\mathrm{3}\:\mathrm{sh}^{\mathrm{2}} \mathrm{t}\:\mathrm{cht}\:\Rightarrow\mathrm{I}\:=\int\:\:\frac{\mathrm{sh}^{\mathrm{2}} \mathrm{t}}{\mathrm{ch}\left(\mathrm{t}\right)}×\mathrm{3sh}^{\mathrm{2}} \mathrm{t}\:\mathrm{cht}\:\mathrm{dt} \\ $$$$=\mathrm{3}\:\int\:\mathrm{sh}^{\mathrm{4}} \mathrm{t}\:\mathrm{dt}\:=\mathrm{3}\:\int\:\left(\frac{\mathrm{ch}\left(\mathrm{2t}\right)−\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \:\mathrm{dt} \\ $$$$=\frac{\mathrm{3}}{\mathrm{4}}\:\int\:\left(\mathrm{ch}^{\mathrm{2}} \left(\mathrm{2t}\right)−\mathrm{2ch}\left(\mathrm{2t}\right)+\mathrm{1}\right)\mathrm{dt} \\ $$$$=\frac{\mathrm{3}}{\mathrm{4}}\:\int\:\mathrm{ch}^{\mathrm{2}} \left(\mathrm{2t}\right)\mathrm{dt}\:−\frac{\mathrm{3}}{\mathrm{2}}\:\int\:\mathrm{ch}\left(\mathrm{2t}\right)\mathrm{dt}\:+\frac{\mathrm{3}}{\mathrm{4}}\mathrm{t} \\ $$$$=\frac{\mathrm{3}}{\mathrm{8}}\int\:\left(\mathrm{1}+\mathrm{ch}\left(\mathrm{4t}\right)\right)\mathrm{dt}\:−\frac{\mathrm{3}}{\mathrm{4}}\mathrm{sh}\left(\mathrm{2t}\right)\:+\frac{\mathrm{3}}{\mathrm{4}}\mathrm{t} \\ $$$$=\frac{\mathrm{3}}{\mathrm{8}}\mathrm{t}\:\:+\frac{\mathrm{3}}{\mathrm{32}}\mathrm{sh}\left(\mathrm{4t}\right)\:−\frac{\mathrm{3}}{\mathrm{4}}\mathrm{sh}\left(\mathrm{2t}\right)+\frac{\mathrm{3}}{\mathrm{4}}\mathrm{t}\:+\mathrm{C} \\ $$$$=\frac{\mathrm{9}}{\mathrm{8}}\mathrm{t}\:+\frac{\mathrm{3}}{\mathrm{16}}\mathrm{sh}\left(\mathrm{2t}\right)\mathrm{ch}\left(\mathrm{2t}\right)−\frac{\mathrm{3}}{\mathrm{2}}\mathrm{sh}\left(\mathrm{t}\right)\mathrm{ch}\left(\mathrm{t}\right)\:+\mathrm{C} \\ $$$$=\frac{\mathrm{9}}{\mathrm{8}}\mathrm{t}\:+\frac{\mathrm{3}}{\mathrm{8}}\mathrm{sh}\left(\mathrm{t}\right)\mathrm{ch}\left(\mathrm{t}\right)\left(\mathrm{2sh}^{\mathrm{2}} \mathrm{t}+\mathrm{1}\right)−\frac{\mathrm{3}}{\mathrm{2}}\mathrm{sh}\left(\mathrm{t}\right)\mathrm{cht}\:+\mathrm{C} \\ $$$$\mathrm{I}=\frac{\mathrm{9}}{\mathrm{8}}\mathrm{ln}\left(\mathrm{x}^{\frac{\mathrm{1}}{\mathrm{3}}} \:+\sqrt{\mathrm{1}+\mathrm{x}^{\frac{\mathrm{2}}{\mathrm{3}}} }\right)\:+\frac{\mathrm{3}}{\mathrm{8}}\:\left(^{\mathrm{3}} \sqrt{\mathrm{x}}\right)\sqrt{\mathrm{1}+\left(^{\mathrm{3}} \sqrt{\mathrm{x}}\right)^{\mathrm{2}} }\left(\mathrm{2}\left(\mathrm{x}\right)^{\frac{\mathrm{2}}{\mathrm{3}}} +\mathrm{1}\right) \\ $$$$−\frac{\mathrm{3}}{\mathrm{2}}\left(^{\mathrm{3}} \sqrt{\mathrm{x}}\right)\sqrt{\mathrm{1}+\left(^{\mathrm{3}} \sqrt{\mathrm{x}}\right)^{\mathrm{2}} }\:+\mathrm{C} \\ $$

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