How to evaluate the integral without using integration by parts












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How can we evaluate the following integral by using substitution rule only?



$$int sqrt{frac{x^2 - 2x}{x^6}},dx$$






calculus integration algebra-precalculus







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  • What have you tried thus far? Context is helpful!
    – Santana Afton
    Nov 21 at 21:11
















0














How can we evaluate the following integral by using substitution rule only?



$$int sqrt{frac{x^2 - 2x}{x^6}},dx$$






calculus integration algebra-precalculus







share|cite|improve this question
























  • What have you tried thus far? Context is helpful!
    – Santana Afton
    Nov 21 at 21:11














0












0








0







How can we evaluate the following integral by using substitution rule only?



$$int sqrt{frac{x^2 - 2x}{x^6}},dx$$






calculus integration algebra-precalculus







share|cite|improve this question















How can we evaluate the following integral by using substitution rule only?



$$int sqrt{frac{x^2 - 2x}{x^6}},dx$$






calculus integration algebra-precalculus




calculus integration algebra-precalculus






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share|cite|improve this question













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edited Nov 21 at 21:42









M. Strochyk

7,67711119




7,67711119










asked Nov 21 at 21:04









Ninja

1,097720




1,097720












  • What have you tried thus far? Context is helpful!
    – Santana Afton
    Nov 21 at 21:11


















  • What have you tried thus far? Context is helpful!
    – Santana Afton
    Nov 21 at 21:11
















What have you tried thus far? Context is helpful!
– Santana Afton
Nov 21 at 21:11




What have you tried thus far? Context is helpful!
– Santana Afton
Nov 21 at 21:11










1 Answer
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Substitute $z=sqrt{dfrac{x-2}{x}}.$ Then
$x-2=xz^2, $ and $$x=dfrac{2}{1-z^2}, x-2 = dfrac{2z^2}{1-z^2}, dx = dfrac{4z}{(1-z^2)^2},dz,$$
$$dfrac{x^2-2x}{x^6}=dfrac{x-2}{x^5} = dfrac{2z^2}{1-z^2} cdot left( dfrac{1-z^2}{2} right)^5 = z^2 cdot left( dfrac{1-z^2}{2} right)^4,$$
therefore,
$$int sqrt{frac{x^2-2x}{x^6}},dx = 4int{z cdot left( dfrac{1-z^2}{2} right)^2 cdot dfrac{z}{(1-z^2)^2},dz} = int{z^2 , dz = ldots}$$






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    1 Answer
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    Substitute $z=sqrt{dfrac{x-2}{x}}.$ Then
    $x-2=xz^2, $ and $$x=dfrac{2}{1-z^2}, x-2 = dfrac{2z^2}{1-z^2}, dx = dfrac{4z}{(1-z^2)^2},dz,$$
    $$dfrac{x^2-2x}{x^6}=dfrac{x-2}{x^5} = dfrac{2z^2}{1-z^2} cdot left( dfrac{1-z^2}{2} right)^5 = z^2 cdot left( dfrac{1-z^2}{2} right)^4,$$
    therefore,
    $$int sqrt{frac{x^2-2x}{x^6}},dx = 4int{z cdot left( dfrac{1-z^2}{2} right)^2 cdot dfrac{z}{(1-z^2)^2},dz} = int{z^2 , dz = ldots}$$






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      Substitute $z=sqrt{dfrac{x-2}{x}}.$ Then
      $x-2=xz^2, $ and $$x=dfrac{2}{1-z^2}, x-2 = dfrac{2z^2}{1-z^2}, dx = dfrac{4z}{(1-z^2)^2},dz,$$
      $$dfrac{x^2-2x}{x^6}=dfrac{x-2}{x^5} = dfrac{2z^2}{1-z^2} cdot left( dfrac{1-z^2}{2} right)^5 = z^2 cdot left( dfrac{1-z^2}{2} right)^4,$$
      therefore,
      $$int sqrt{frac{x^2-2x}{x^6}},dx = 4int{z cdot left( dfrac{1-z^2}{2} right)^2 cdot dfrac{z}{(1-z^2)^2},dz} = int{z^2 , dz = ldots}$$






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        Substitute $z=sqrt{dfrac{x-2}{x}}.$ Then
        $x-2=xz^2, $ and $$x=dfrac{2}{1-z^2}, x-2 = dfrac{2z^2}{1-z^2}, dx = dfrac{4z}{(1-z^2)^2},dz,$$
        $$dfrac{x^2-2x}{x^6}=dfrac{x-2}{x^5} = dfrac{2z^2}{1-z^2} cdot left( dfrac{1-z^2}{2} right)^5 = z^2 cdot left( dfrac{1-z^2}{2} right)^4,$$
        therefore,
        $$int sqrt{frac{x^2-2x}{x^6}},dx = 4int{z cdot left( dfrac{1-z^2}{2} right)^2 cdot dfrac{z}{(1-z^2)^2},dz} = int{z^2 , dz = ldots}$$






        share|cite|improve this answer












        Substitute $z=sqrt{dfrac{x-2}{x}}.$ Then
        $x-2=xz^2, $ and $$x=dfrac{2}{1-z^2}, x-2 = dfrac{2z^2}{1-z^2}, dx = dfrac{4z}{(1-z^2)^2},dz,$$
        $$dfrac{x^2-2x}{x^6}=dfrac{x-2}{x^5} = dfrac{2z^2}{1-z^2} cdot left( dfrac{1-z^2}{2} right)^5 = z^2 cdot left( dfrac{1-z^2}{2} right)^4,$$
        therefore,
        $$int sqrt{frac{x^2-2x}{x^6}},dx = 4int{z cdot left( dfrac{1-z^2}{2} right)^2 cdot dfrac{z}{(1-z^2)^2},dz} = int{z^2 , dz = ldots}$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 21 at 21:41









        M. Strochyk

        7,67711119




        7,67711119






























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