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Integral 1/(sqroot(84+16x+4x^2))

  1. May 20, 2014 #1
    Could you please check this integration?

    I get a different answer from WolframAlpha.

    Integral of 1/(sqroot(84+16x+4x^2)) dx

    Look at picture please! I apologize for the other physiology information. I have no more clean scrap paper left.
     

    Attached Files:

  2. jcsd
  3. May 20, 2014 #2

    pasmith

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    Homework Helper

    So far as [tex]
    \int \frac 1{\sqrt{84 + 16x + 4x^2}}\,dx = \frac14 \int \frac{2(x+2)}{(x + 2)\sqrt{(x+ 2)^2 + 17}}\,dx[/tex] I agree with your calculations. You now appear to substitute [itex]u = (x + 2)^2 + 17[/itex], with [itex]du = 2(x + 2)\,dx[/itex], from which you should obtain [tex]
    \frac14 \int \frac{2(x+2)}{(x + 2)\sqrt{(x+ 2)^2 + 17}}\,dx = \frac14 \int \frac{1}{(x + 2)\sqrt{u}}\,du = \frac14 \int \frac{1}{\sqrt{u - 17}\sqrt{u}}\,du[/tex] rather than [tex]\frac14 \int \frac{1}{u\sqrt{u^2 + 17}}\,du[/tex] which is what you have.

    It would have been better to substitute [itex]x + 2 = \sqrt{17} \sinh \theta[/itex] after completing the square in the denominator. The motivation for this is that [itex]\cosh^2 \theta = 1 + \sinh^2 \theta[/itex] and the derivative of [itex]\sinh \theta[/itex] is [itex]\cosh \theta[/itex]. You therefore end up with a constant integrand.
     
  4. May 20, 2014 #3
    The issue here is when you do the u substitution and plug in your values for u. You should have written

    ##\frac{1}{4}\displaystyle\int\frac{1}{\sqrt{u-17}\sqrt{u}}du##


    I wouldn't do u-substitution for this problem. Try doing trig substitution: ##x+2=\sqrt{17}\tan{\theta}##
     
  5. May 20, 2014 #4
    Hi, I have never been introduced to those wild "sinh" functions! Let me get back to you after a few minutes of sweat.
     
  6. May 20, 2014 #5
    Oh. I understand. Then, I'm in a bit of a sticky situation... Since, I cannot use sinh cosh etc.., I'm stuck with a trig sub.
     
  7. May 20, 2014 #6
    Hmm. At which step should I do the trig sub?
     
  8. May 20, 2014 #7
    The step after you complete the square in the denominator. If you haven't learned hyberbolic trig functions, then use my suggestion. I have a feeling pasmith's substitution is a bit cleaner but there is nothing incorrect about using tangent.
     
  9. May 20, 2014 #8
    Dear Sirs/Madams,
    Here is my second attempt. But, I graphed it and it does not give the same answer as the hyperbolic function that wolfram alpha gives me.
     

    Attached Files:

  10. May 20, 2014 #9
    That looks good to me. What did Wolfram give as an answer? Likely, you could manipulate that answer into the answer you gave.
     
  11. May 20, 2014 #10
    ArcSinh[(2 + x)/Sqrt[17]]/2
     
  12. May 20, 2014 #11
    The functions do not match!
     
  13. May 20, 2014 #12
    They are equivalent to each other. Plug their difference into WolframAlpha to see:

    ArcSinh[(2 + x)/Sqrt[17]]/2 - 1/2log(sqrt{(x+2)^2+17}/sqrt(17)+(x+2)/sqrt(17))

    A good exercise for you would be to use the definition of arcsinh to write that form into your form.

    As a starting point: ##\sinh^{-1}{x}=\ln{(x+\sqrt{1+x^2})}##
     
  14. May 20, 2014 #13

    Ray Vickson

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    If you set ##y = x+2## you have an integral of the form
    [tex] \int \frac{dy}{\sqrt{y^2 + a^2}} \: (a > 0) [/tex]
    (times a constant), and if you further put ##y = \sqrt{a}\: u## you end up having to do the integral
    [tex]
    \int \frac{du}{\sqrt{u^2+1}} [/tex]
    You might as well do this integral once-and-for-all, then use it in your current problem.
     
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