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2 integrals - - (within next 30 mins if possible)

  1. Feb 26, 2008 #1
    2 integrals - URGENT - (within next 30 mins if possible)

    1. The problem statement, all variables and given/known data

    [tex]\int_0^{b}\frac{dx}{x+\sqrt x}[/tex]

    [tex]\int[/tex][tex]^{b}_{}[/tex] (x + [tex]\sqrt{x})^{-1}[/tex]dx
    [tex]_{ 1}[/tex]

    3. The attempt at a solution

    For the first i have tried x=t^2 but ended up with something i cant integrate (2t/(t^2 + t) dt

    for the second i tried u= 1 + x^3 but that did'nt help -
     
    Last edited: Feb 26, 2008
  2. jcsd
  3. Feb 26, 2008 #2

    morphism

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    Hint:
    [tex]\frac{2t}{t^2 + t} = \frac{2t+1}{t^2 + t} - \frac{1}{t^2 + t}[/tex]

    This screams substitution for the first term and partial fractions for the second.

    Your second integrand is exactly like your first one. Am I missing something?
     
  4. Feb 26, 2008 #3
    ahhhh- yeah the second is

    [tex]\int[/tex][tex]^{\infty}_{}[/tex] (1 + x[tex]^{3}[/tex])[tex]^{-1/2}[/tex] dx
    [tex]_{ 1}[/tex]
     
  5. Feb 26, 2008 #4

    olgranpappy

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    try the substitution

    y=sqrt[x]

    ...oh, right... that is what you tried... then factor out a y from numerator and denominator...
     
  6. Feb 26, 2008 #5
    cool - cheers
     
  7. Feb 26, 2008 #6
    [tex]\int_0^{b}\frac{dx}{x+\sqrt x}[/tex]

    This thing works out nicely if you factor out [tex]\sqrt{x}[/tex], giving you

    [tex]\sqrt{x}\left(\sqrt{x} + 1\right)[/tex]

    Then let u = [tex]\left(\sqrt{x} + 1\right)[/tex] and you're almost done.
     
  8. Feb 26, 2008 #7
  9. Feb 26, 2008 #8
    yeah - i got the x= t^2 but i could't integrate (2t/(t^2 + t) dt

    thanks though
     
  10. Feb 26, 2008 #9
    i also simplified it in there, now you will end up with 2 ln(1+t) on the interval sqrt b, and 0
     
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