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Integration with power substitution

  1. Mar 29, 2016 #1
    1. The problem statement, all variables and given/known data
    Find the integral [itex]\int \frac{2\sqrt[5]{2x-3}-1}{(2x-3)\sqrt[5]{2x-3}+\sqrt[5]{2x-3}}\mathrm dx[/itex]

    2. The attempt at a solution
    [tex]\frac{2\sqrt[5]{2x-3}-1}{(2x-3)\sqrt[5]{2x-3}+\sqrt[5]{2x-3}}=\frac{2\sqrt[5]{2x-3}-1}{2\sqrt[5]{2x-3}(x-1)}=\frac{1}{x-1}-\frac{1}{2\sqrt[5]{2x-3}(x-1)}[/tex]


    [tex]\int \frac{1}{x-1}\mathrm dx=\ln|x-1|+c[/tex]
    [tex]\int \frac{1}{2\sqrt[5]{2x-3}(x-1)}\mathrm dx=\frac{1}{2}\int \frac{1}{\sqrt[5]{2x-3}(x-1)}\mathrm dx[/tex]

    Substitution [itex]u=2x-3,du=2dx[/itex] gives
    [tex]\frac{1}{2}\int \frac{1}{\sqrt[5]{2x-3}(x-1)}\mathrm dx=\int \frac{1}{u^{6/5}+u^{1/5}}\mathrm du[/tex]

    Substitution [itex]u^{1/5}=v,u=v^5,du=5v^4[/itex] gives
    [tex]\int \frac{1}{u^{6/5}+u^{1/5}}\mathrm du=5\int \frac{v^3}{v^5+1}\mathrm dv[/tex]

    [tex]v^5+1=(v+1)(v^4-v^3+v^2-v+1)[/tex]

    Using partial fractions:
    [tex]\frac{v^3}{v^5+1}=\frac{A}{v+1}+\frac{Bv^3+Cv^2+Dv+E}{v^4-v^3+v^2-v+1}[/tex]

    [tex]\Rightarrow A=-1/3,B=1/3,C=1,D=-2/3,E=1/3[/tex]

    [tex]\int \frac{A}{v+1}\mathrm dv=-\frac{1}{3}\ln|v+1|+c[/tex]

    How to integrate [tex]\frac{Bv^3+Cv^2+Dv+E}{v^4-v^3+v^2-v+1}?[/tex]
     
  2. jcsd
  3. Mar 29, 2016 #2

    Ssnow

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    There is an other half when you substitute ##2x-3=u## because ##dx=\frac{du}{2}##, but this not the principal problem ... other things seems correct, for your last integral you must factorize the polynomial ##v^4-v^{3}+v^{2}-v+1## that has four solution in ##\mathbb{C}## and proceed with the partial fractions in the complex case ...
     
  4. Mar 29, 2016 #3
    I found the easier method to integrate [itex]\int \frac{1}{\sqrt[5]{2x-3}(x-1)}\mathrm dx[/itex].
    [tex]\int \frac{1}{\sqrt[5]{2x-3}(x-1)}\mathrm dx=\int (x-1)^{-1}(-3+2x)^{-1/5}\mathrm dx[/tex]
    which is the integral of differential binomial.

    After substitution [itex]u=x-1,du=dt[/itex],
    [tex]\int u^{-1}(-1+2u)^{-1/5}\mathrm du[/tex]

    which can be solved by substitution [itex]-1+2u=v^5,du=\frac{5v^4}{2}dv[/itex]
    [tex]\Rightarrow \int \frac{1}{\sqrt[5]{2x-3}(x-1)}\mathrm dx=10\int \frac{v^4}{v^5+1}\mathrm dv[/tex]
     
  5. Mar 29, 2016 #4

    Samy_A

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    Didn't you loose the ##v## from ##(-1+2u)^{-1/5}## in the denominator?
     
  6. Mar 29, 2016 #5
    You are right, it should be [itex]10\int \frac{v^4}{(v^5+1)^2}\mathrm dv[/itex].
     
  7. Mar 29, 2016 #6

    Samy_A

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    No. What is ##(-1+2u)^{-1/5}## in terms of ##v##, if ##-1+2u=v^5##?
     
  8. Mar 29, 2016 #7
    [itex](-1+2u)^{-1/5}=v^{-1}\Rightarrow 5\int \frac{v^3}{v^5+1}\mathrm dv[/itex]
     
  9. Mar 29, 2016 #8

    Samy_A

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    You again lost a factor 2, as @Ssnow noticed.
    I think the correct integral is ##10 \int \frac{v^3}{v^5+1}\mathrm dv##
     
  10. Mar 29, 2016 #9
    Thanks for the help.
     
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