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Antiderivative involving a radical

  1. Nov 8, 2014 #1
    1. The problem statement, all variables and given/known data
    [tex]\int\frac{1}{(x+1)^2}\sqrt{\frac{x}{1-x}}{\rm{d}}x[/tex]

    2. Relevant equations


    3. The attempt at a solution
    Utterly perplexed. Have no ideas how to do this one. Did try bringing the entire thing under square root and try partial fractions, but the entire thing is modified by the square root. I can't integrate the sum separately or anything.
    Integration by parts is even more devastating and more incomprehensible. Not sure how to proceed. Is there a cunning substitution to be made? Hints, please.
    Thanks
     
  2. jcsd
  3. Nov 8, 2014 #2

    SteamKing

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    Partial fractions aren't recommended because of the square root.

    Since you have various powers of linear terms in x, have you tried u-substitution?
     
  4. Nov 11, 2014 #3
    Alright, I managed this so far:
    substituting the radical for u:
    [tex]u = \sqrt{\frac{x}{1-x}} \Rightarrow x = \frac{u^2}{u^2 +1}, {\rm{d}}x = \frac{2u}{(u^2 +1)^2}{\rm{d}}u[/tex]
    resulting in
    [tex]\int\frac{2u^2}{(2u^2 + 1)^2}{\rm{d}}u[/tex]
    adding 1 and -1 in the numerator:
    [tex]\int\frac{{\rm{d}}u}{2u^2 + 1} - \int\frac{{\rm{d}}u}{(2u^2 + 1)^2}[/tex]
    The first one is ok, but the second integral is troublesome. Partial fractions yield no result.
     
  5. Nov 11, 2014 #4

    SteamKing

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    I would have tried something like u = 1 + x or u = 1 - x, rather than mess with the radical as a whole.
     
  6. Nov 11, 2014 #5
    My thought process went along the lines of "get rid of the square root, somehow". How will I be able to integrate the second integral I wound up with? I feel like I'm on the right track.
    EDIT: Right as I posted, I thought about trigonometric substitution. To be continued..
     
  7. Nov 11, 2014 #6

    HallsofIvy

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    I don't know why you say "partial fractions" doesn't work. They usual "partial fractions" format says that we can write this as
    [tex]\frac{1}{(u^2+ 1)^2}= \frac{Au+ B}{u^2+ 1}+ \frac{Cu+ D}{(u^2+ 1)^2}[/tex]
    There are a number of different ways to solve for A, B, C, and D. Perhaps the simplest is to multiply both sides by [tex](u^2+ 1)^2[/tex] to get
    [tex]1= (Au+ B)(u^2+ 1)+ Cu+ D[/tex]
    This must be true for all u. Setting u to 4 different values will give 4 different equations to solve for A, B, C, and D.
    Setting u= 0, B+ D= 1.
    Setting u= 1, 2A+ 2B+ C+ D= 1.
    Setting u= -1, -2A+2B- 2C+ D= 1.
    Setting u= 2, 10A+ 5B+ 2C+ D= 1.

    Solve those four equations for A, B, C, and D. I would start by adding the last two equation to get an equation with only A, B, and D in it, then Add twice the second equation to get another equation with only A, B, and D. Combine those two equations to eliminate A. Combine that equation, with only B and D, with B+ D= 1, to eliminate either B or D.
     
  8. Nov 11, 2014 #7
    Managed to figure it out: In the second integral i can substitute
    [tex]\tan{t} = \sqrt{2}u , \frac{{\rm{d}}t}{\cos^2{t}} = \sqrt{2}{\rm{d}}u\\ \frac{1}{\sqrt{2}}\int\frac{{\rm{d}}t}{(\tan^2{t}+1)^2 \cos^2{t}}[/tex]
    [itex]\tan^2{x} +1 = \frac{1}{\cos^2{x}}[/itex] so we get that it is:
    [tex]\frac{1}{\sqrt{2}}\int \cos^2{t}{\rm{d}}t[/tex]which is a piece of cake.

    Thank you HallsofIvy, I will try your suggestion aswell, I must have done something wrong when I tried partial fractions.
     
  9. Nov 11, 2014 #8

    LCKurtz

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    That doesn't seem likely to get anywhere since A=B=C=0, D=1 is the original problem.
     
  10. Nov 11, 2014 #9

    pasmith

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    You can now factor the integrand as [tex]
    \int u \frac{2u}{(2u^2 + 1)^2}\,du[/tex] and integrate by parts.
     
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