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dx-x-1-3-x-2-5-1-4-




Question Number 159315 by cortano last updated on 15/Nov/21
 ∫ (dx/( (((x−1)^3 (x+2)^5 ))^(1/4)  )) ?
$$\:\int\:\frac{{dx}}{\:\sqrt[{\mathrm{4}}]{\left({x}−\mathrm{1}\right)^{\mathrm{3}} \left({x}+\mathrm{2}\right)^{\mathrm{5}} }\:}\:? \\ $$
Commented by tounghoungko last updated on 15/Nov/21
Y=∫ (dx/((x+2)^2 (((x−1)/(x+2)))^(3/4) ))   let (((x−1)/(x+2)))^(1/4)  = v ⇒x=((2v^4 +1)/(1−v^4 ))  Y= ∫(1/((((2v^4 +1+2−2v^4 )/(1−v^4 )))^2 v^3 )) (((12v^3 )/((1−v^4 )^2 )))dv  Y= ∫ ((12)/9) dv = (4/3)v +c = (4/3)(((x−1)/(x+2)))^(1/4)  +c
$${Y}=\int\:\frac{{dx}}{\left({x}+\mathrm{2}\right)^{\mathrm{2}} \left(\frac{{x}−\mathrm{1}}{{x}+\mathrm{2}}\right)^{\mathrm{3}/\mathrm{4}} }\: \\ $$$${let}\:\sqrt[{\mathrm{4}}]{\frac{{x}−\mathrm{1}}{{x}+\mathrm{2}}}\:=\:{v}\:\Rightarrow{x}=\frac{\mathrm{2}{v}^{\mathrm{4}} +\mathrm{1}}{\mathrm{1}−{v}^{\mathrm{4}} } \\ $$$${Y}=\:\int\frac{\mathrm{1}}{\left(\frac{\mathrm{2}{v}^{\mathrm{4}} +\mathrm{1}+\mathrm{2}−\mathrm{2}{v}^{\mathrm{4}} }{\mathrm{1}−{v}^{\mathrm{4}} }\right)^{\mathrm{2}} {v}^{\mathrm{3}} }\:\left(\frac{\mathrm{12}{v}^{\mathrm{3}} }{\left(\mathrm{1}−{v}^{\mathrm{4}} \right)^{\mathrm{2}} }\right){dv} \\ $$$${Y}=\:\int\:\frac{\mathrm{12}}{\mathrm{9}}\:{dv}\:=\:\frac{\mathrm{4}}{\mathrm{3}}{v}\:+{c}\:=\:\frac{\mathrm{4}}{\mathrm{3}}\sqrt[{\mathrm{4}}]{\frac{{x}−\mathrm{1}}{{x}+\mathrm{2}}}\:+{c}\: \\ $$

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