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Intergal of x^2/sqrt(x^2+3)

  1. Sep 6, 2011 #1
    1. The problem statement, all variables and given/known data

    [itex]\int \frac{x^2}{\sqrt{x^2+3}}[/itex]

    2. Relevant equations

    sinh-1(u) = u' / (u^2 + 1)

    3. The attempt at a solution

    Make the x^2 + 3 look like x^2 + 1 by taking out a sqrt(3). Giving you

    [itex]\int \frac{x^2}{\sqrt{3} \sqrt{\frac{x^2}{3}+1}}[/itex]

    Set the constant outside the integral.

    [itex] \frac{1}{\sqrt{3}} \int \frac{x^2}{\sqrt{\frac{x^2}{3}+1}}[/itex]

    Now we find where [itex]u^2 = \frac{x^2}{3} [/itex] , which is [itex]u = \frac{x}{\sqrt{3}} [/itex]. Now we know the u of the sinh-1, we find u'

    [itex]u' = \frac{1}{\sqrt{3}} [/itex]


    So now we taken care of everything but x^2...

    Where to go now?
     
  2. jcsd
  3. Sep 6, 2011 #2

    micromass

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    So, now substite in the sinh. You'll see that it works out fine...

    What do you get??
     
  4. Sep 6, 2011 #3
    I know that the sinh^-1(x/sqrt(3)) is in the answer, but there is still a multiplication between x^2 and the the sin so more needs to be done.

    I can't just have

    (1/3)x^3 * sinh^-1(x/sqrt(3)) + c

    because the product rule states uv' + u'v, if it was as easy as u'v' then it would work.
     
  5. Sep 6, 2011 #4

    micromass

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    :confused: You have the integral

    [tex]\int{\frac{x^2}{\sqrt{x^2+1}}dx}[/tex]

    after your substitutions right?? So just substitute in [itex]x=sinh(u)[/itex]
     
  6. Sep 6, 2011 #5

    HallsofIvy

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    Micromass likes his hyperbolic functions substitutions! Perhaps just because i learned them first, I always think of trig substitutions first. Here, after taking out a [itex]\sqrt{3}[/itex], you have [itex]\sqrt{1+(x/\sqrt{3})^2}[/itex] and since [itex]1+ tan^2(\theta)= sec^2(\theta)[/itex] I would let [itex]x= \sqrt{3}tan(\theta)[/itex].

    Of course, then, the [itex]x^2[/itex] in the numerator becomes [itex]3tan^2(\theta)[/itex]
     
  7. Sep 6, 2011 #6
    [itex]\sqrt{1+(x/\sqrt{3})^2}[/itex]

    Shouldnt that be 3 not sqrt(3)

    I still don't get it. I need more of a step by step. coming to the solution wolfgram comes to:

    Image%202011-09-06%20at%208.06.47%20PM.png

    I see how you decided what x is, but I dont see how they decided what u was.
     
  8. Sep 6, 2011 #7

    HallsofIvy

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    No, it shouldn't. Notice the parentheses. Both x and [itex]\sqrt{3}[/itex] are squared.

    The crucial point is that you have [itex]\sqrt{1+ a^2}[/itex] and want to get rid of the square root. You should immediately think of the trig identity [itex]1+ tan^2\theta= sec^2\theta[/itex].
     
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