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Integral of 1/(sqrt(x)-x)

  1. Oct 7, 2007 #1
    1. The problem statement, all variables and given/known data
    integrate 1/(sqrt(x)-x)

    2. Relevant equations

    3. The attempt at a solution

    1st I tried to do it the "regular" way, I know that it must be the derivative of

    ln(sqrt(x)-x), but the derivative of sqrt(x)-x is not 1. So I added stuff to the equation to make it like 1/(sqrt(x)-x) = ((1/2)(x)^(-1/2)-(1/2)(x)^(-1/2)-1+2)/(sqrt(x)-x)... and more things goes on and I got something like (4*sqrt(x)-1)/(2*(x-x*sqrt(x)))... which is obviously not the answer.

    2nd I tried to use t = sqrt(x) and 1/(sqrt(x)-x) = dt^2/(t-t^2), but I have no idea how to solve this....
  2. jcsd
  3. Oct 8, 2007 #2
    is your problem?

  4. Oct 8, 2007 #3
  5. Oct 8, 2007 #4


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    If t=sqrt(x), then dt/dx=1/(2sqrt(x))=1/(2t). So your integrand should become 2t/(t-t^2). It should be pretty easy to take it from here.
  6. Oct 8, 2007 #5


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    When you have the square root of something in an integral, it's offen helpful to make the substitution u2 = thing that's under the square root sign.
  7. Oct 8, 2007 #6
    is the answer :

    2* ln(1-x^2) +c
  8. Oct 8, 2007 #7


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    transgalactic = beanryu?

    And did you try differentiating it and check what comes out?
  9. Oct 8, 2007 #8
    i tried the method of

    it gives me that answer

    2* ln(1-x^2) +c

    beanryu is it the right answer??
    Last edited: Oct 8, 2007
  10. Oct 8, 2007 #9
    Take the derivative and if you get the same thing that is in your integral, you are right.
  11. Oct 8, 2007 #10
    unfortunatly no

    but where i did wrong

    { is integral sign

    made x=v^2 >>> dx=2v*dv

    then it gives us

    { (2v*dv)/(v-v^2) that gives us

    { 2*dv/(1-v) = 2*ln(1-v) = 2*ln(1-x^2)
    Last edited: Oct 8, 2007
  12. Oct 8, 2007 #11


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    1) The integral of dv/(1-v) is NOT ln(1-v). (Close, but not quite. Hint: if you define u=v-1, what do you get?)

    2) v is NOT equal to x^2.
  13. Oct 8, 2007 #12
    its -ln(1-v)

    what do meen v is not equal to x^2
    i defined that
  14. Oct 8, 2007 #13


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    No, you defined v^2 = x.
    What does that make v?

    (By the way, you're correct on the -ln(1-v) now)
  15. Oct 8, 2007 #14
    damn little mistakes
  16. Oct 8, 2007 #15
    if x=v^2 >>> shouldn't dx = d(v^2)?
  17. Oct 8, 2007 #16
    i got the answer,
    my answer is (-2)ln(sqrt(x)-1) using u^2 = x method,

    than the answer in the textbook is (-2)ln(1-sqrt(x)),

    I took both derivativea they are the same.... wierd....

    if dx/(sqrt(x)-x) = dy/(tan(y)), should I use answer (-2)*ln(1-sqrt(x)), because this is the function of x when x<1.

    thank you guys~
    Last edited: Oct 8, 2007
  18. Oct 9, 2007 #17
    dx = d(v^2)=2v*dv
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