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Question Number 111017 by bemath last updated on 01/Sep/20
(√(bemath)) ∫ (dx/( ((4−((3−2x))^(1/(3 )) ))^(1/(4 )) )) ?
$$\:\:\:\:\sqrt{\mathrm{bemath}} \\ $$$$\int\:\frac{\mathrm{dx}}{\:\sqrt[{\mathrm{4}\:}]{\mathrm{4}−\sqrt[{\mathrm{3}\:}]{\mathrm{3}−\mathrm{2x}}}}\:? \\ $$
Answered by john santu last updated on 01/Sep/20
by letting ν = ((4−((3−2x))^(1/(3 )) ))^(1/(4 )) ⇒ν^4 =4−((3−2x))^(1/(3 )) ⇒3−2x = (4−ν^4 )^3 2x = 3−(4−ν^4 )^3 ⇒2dx = −3.(−4ν^3 )(4−ν^4 )^2 dν I=∫ ((12ν^3 (4−ν^4 )^2 )/(2ν)) dν I=6∫ν^2 (16−8ν^4 +ν^8 ) dν I=6∫(16ν^2 −8ν^6 +ν^(10) )dν I=6(((16ν^3 )/3)−(8/7)ν^7 +(1/(11))ν^(11) )+c I=6(((4−((3−2x))^(1/(3 )) )^3 ))^(1/(4 )) (((16)/3)−(8/7)(4−((3−2x))^(1/(3 )) )+(1/(11))(4−((3−2x))^(1/(3 )) )^2 )+ c
$${by}\:{letting}\:\nu\:=\:\sqrt[{\mathrm{4}\:}]{\mathrm{4}−\sqrt[{\mathrm{3}\:}]{\mathrm{3}−\mathrm{2}{x}}} \\ $$$$\Rightarrow\nu^{\mathrm{4}} =\mathrm{4}−\sqrt[{\mathrm{3}\:}]{\mathrm{3}−\mathrm{2}{x}}\:\Rightarrow\mathrm{3}−\mathrm{2}{x}\:=\:\left(\mathrm{4}−\nu^{\mathrm{4}} \right)^{\mathrm{3}} \\ $$$$\mathrm{2}{x}\:=\:\mathrm{3}−\left(\mathrm{4}−\nu^{\mathrm{4}} \right)^{\mathrm{3}} \\ $$$$\Rightarrow\mathrm{2}{dx}\:=\:−\mathrm{3}.\left(−\mathrm{4}\nu^{\mathrm{3}} \right)\left(\mathrm{4}−\nu^{\mathrm{4}} \right)^{\mathrm{2}} \:{d}\nu \\ $$$${I}=\int\:\frac{\mathrm{12}\nu^{\mathrm{3}} \left(\mathrm{4}−\nu^{\mathrm{4}} \right)^{\mathrm{2}} }{\mathrm{2}\nu}\:{d}\nu \\ $$$${I}=\mathrm{6}\int\nu^{\mathrm{2}} \left(\mathrm{16}−\mathrm{8}\nu^{\mathrm{4}} +\nu^{\mathrm{8}} \right)\:{d}\nu \\ $$$${I}=\mathrm{6}\int\left(\mathrm{16}\nu^{\mathrm{2}} −\mathrm{8}\nu^{\mathrm{6}} +\nu^{\mathrm{10}} \right){d}\nu \\ $$$${I}=\mathrm{6}\left(\frac{\mathrm{16}\nu^{\mathrm{3}} }{\mathrm{3}}−\frac{\mathrm{8}}{\mathrm{7}}\nu^{\mathrm{7}} +\frac{\mathrm{1}}{\mathrm{11}}\nu^{\mathrm{11}} \right)+{c} \\ $$$${I}=\mathrm{6}\sqrt[{\mathrm{4}\:}]{\left(\mathrm{4}−\sqrt[{\mathrm{3}\:}]{\mathrm{3}−\mathrm{2}{x}}\right)^{\mathrm{3}} }\:\left(\frac{\mathrm{16}}{\mathrm{3}}−\frac{\mathrm{8}}{\mathrm{7}}\left(\mathrm{4}−\sqrt[{\mathrm{3}\:}]{\mathrm{3}−\mathrm{2}{x}}\right)+\frac{\mathrm{1}}{\mathrm{11}}\left(\mathrm{4}−\sqrt[{\mathrm{3}\:}]{\mathrm{3}−\mathrm{2}{x}}\right)^{\mathrm{2}} \:\right)+\:{c}\: \\ $$
Answered by Dwaipayan Shikari last updated on 01/Sep/20
∫(dx/( ((4−a))^(1/4) )) 3−2x=a^3 −2=3a^2 (da/dx) −(1/2)∫((3a^2 da)/((4−a)^(1/4) )) (4−a=t^4 ⇒−1=4t^3 (dt/da) (3/2)∫((a^2 4t^3 dt)/t)=6∫t^2 (4−t^4 )^2 dt=6∫16t^2 −8t^6 +t^(10) =32t^3 −((48)/7)t^7 +(6/(11))t^(11) +C =32(4−a)^(3/4) −((48)/7)(4−a)^(7/4) +(6/(11))(4−a)^((11)/4) +C =32(4−((3−2x))^(1/3) )^(3/4) −((48)/7)(4−((3−2x))^(1/3) )^(7/4) +(6/(11))(4−((3−2x))^(1/3) )^((11)/4) +C
$$\int\frac{{dx}}{\:\sqrt[{\mathrm{4}}]{\mathrm{4}−{a}}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{3}−\mathrm{2}{x}={a}^{\mathrm{3}} \:\:\:\:−\mathrm{2}=\mathrm{3}{a}^{\mathrm{2}} \frac{{da}}{{dx}}\:\:\:\:\:\:\:\: \\ $$$$−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{3}{a}^{\mathrm{2}} {da}}{\left(\mathrm{4}−{a}\right)^{\frac{\mathrm{1}}{\mathrm{4}}} }\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{4}−{a}={t}^{\mathrm{4}} \Rightarrow−\mathrm{1}=\mathrm{4}{t}^{\mathrm{3}} \frac{{dt}}{{da}}\right. \\ $$$$\frac{\mathrm{3}}{\mathrm{2}}\int\frac{{a}^{\mathrm{2}} \mathrm{4}{t}^{\mathrm{3}} {dt}}{{t}}=\mathrm{6}\int{t}^{\mathrm{2}} \left(\mathrm{4}−{t}^{\mathrm{4}} \right)^{\mathrm{2}} {dt}=\mathrm{6}\int\mathrm{16}{t}^{\mathrm{2}} −\mathrm{8}{t}^{\mathrm{6}} +{t}^{\mathrm{10}} \\ $$$$=\mathrm{32}{t}^{\mathrm{3}} −\frac{\mathrm{48}}{\mathrm{7}}{t}^{\mathrm{7}} +\frac{\mathrm{6}}{\mathrm{11}}{t}^{\mathrm{11}} +{C} \\ $$$$=\mathrm{32}\left(\mathrm{4}−{a}\right)^{\frac{\mathrm{3}}{\mathrm{4}}} −\frac{\mathrm{48}}{\mathrm{7}}\left(\mathrm{4}−{a}\right)^{\frac{\mathrm{7}}{\mathrm{4}}} +\frac{\mathrm{6}}{\mathrm{11}}\left(\mathrm{4}−{a}\right)^{\frac{\mathrm{11}}{\mathrm{4}}} +{C} \\ $$$$=\mathrm{32}\left(\mathrm{4}−\sqrt[{\mathrm{3}}]{\mathrm{3}−\mathrm{2}{x}}\right)^{\frac{\mathrm{3}}{\mathrm{4}}} −\frac{\mathrm{48}}{\mathrm{7}}\left(\mathrm{4}−\sqrt[{\mathrm{3}}]{\mathrm{3}−\mathrm{2}{x}}\right)^{\frac{\mathrm{7}}{\mathrm{4}}} +\frac{\mathrm{6}}{\mathrm{11}}\left(\mathrm{4}−\sqrt[{\mathrm{3}}]{\mathrm{3}−\mathrm{2}{x}}\right)^{\frac{\mathrm{11}}{\mathrm{4}}} +{C} \\ $$

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