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Antiderivative for 1/[x²√(1+x²)]?

  1. May 19, 2005 #1
    The title sums it up. How can I find this?

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

    I've tried a lot of different things to come up with an answer. My answers must be wrong, but I don't see where I'm making a mistake. Here's one way I've tried to solve the problem:

    Let
    [itex]u = (1+x^2)^{-1/2}[/itex]
    [itex]v = -x^{-1}[/itex]

    Then
    [itex]u' = -x(1+x^2)^{-3/2}[/itex]
    [itex]v' = x^{-2}[/itex]

    And so the original problem is equivalent to:

    [tex]\int uv' = uv - \int u'v[/tex]
    [tex]= -\frac {1}{x (1+x^2)^{1/2}} - \int \frac{(-x)(-x^{-1}) dx}{(1+x^2)^{3/2}}[/tex]
    [tex]= -\frac {1}{x (1+x^2)^{1/2}} - \int \frac{dx}{(1+x^2)^{3/2}}[/tex]

    ... which calls for this trigonometric substitution:

    tan y = x
    [tex]\frac{dy}{\cos^2 y} = dx[/tex]

    So, with that substitution, the original problem is equal to this:

    [tex]-\frac {1}{x (1+x^2)^{1/2}} - \int \frac{dy}{\cos^2 y(\frac{1}{\cos^2 y})^{3/2}}[/tex]
    [tex]= -\frac {1}{x (1+x^2)^{1/2}} - \int \cos y \; dy[/tex]

    which is

    [tex]= -\frac {1}{x (1+x^2)^{1/2}} - \sin \arctan x + C[/tex]

    The book's answer doesn't have any trigonometric functions in it, and more to the point, differentiating my answer doesn't give me the function inside the original problem, so that's out.

    Where did I go wrong?
     
  2. jcsd
  3. May 19, 2005 #2
    [tex]\int \frac{dx}{x^2 \sqrt{1+x^2}}[/tex]

    Try the trig substitution in the beginning.
    [tex] x = tan u, dx = sec^2udu[/tex]

    [tex] \int\frac{sec^2u}{\tan^2usecu}du = \int \frac{secu}{\tan^2u}du = \int \frac{1}{\cos u \tan^2u}du = \int \frac{1}{\frac{sin^2u}{cos^2u}cosu}du = \int \frac{cos u}{sin^2u}du[/tex]

    A nice substitution will take care of that, and when your done with the simplifying you should have a trig function. Draw a triangle and figure out the final result which should be

    [tex]\frac{-\sqrt{x^2+1}}{x} [/tex]
     
    Last edited: May 19, 2005
  4. May 19, 2005 #3

    arildno

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    Note that:
    [tex]x=tan(arctan(x))=\frac{\sin(arctan(x))}{\sqrt{1-\sin^{2}(arctan(x))}}[/tex]
    That is, we get:
    [tex]sin(arctan(x))=\frac{x}{\sqrt{1+x^{2}}}[/tex]

    Try and see if this is what you needed..
     
  5. May 19, 2005 #4

    arildno

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    Note that your own answer agrees with whozum's by using the expression provided for sin(arctan(x))
     
  6. May 19, 2005 #5
    Mine is simpler!!! :)
     
  7. May 19, 2005 #6

    arildno

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    Sure enough.
    Personally, however, I would have used the substitution x=Sinh(u) from the start, rather than a trigonometric substitution.
     
  8. May 19, 2005 #7
    I dont know hyperbolic trigonometry, can you believe Im a physics junior and have never been taught it?
     
  9. May 19, 2005 #8
    Thanks SO much! I thought there might be something I could substitute for sin atan x, but I didn't figure out what it should be.
     
  10. May 19, 2005 #9

    arildno

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    You'll soon learn it; it isn't that difficult.
     
  11. May 19, 2005 #10

    arildno

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    Have you tried to follow whozum's suggestion of drawing a diagram and see how this identity follows:
    [tex]\sin(arctan(x))=\frac{x}{\sqrt{1+x^{2}}}[/tex]
    If you haven't, you should try it when you find the time to do so.
     
  12. May 19, 2005 #11
    I think, that substitution t=1/x isn't more difficult.
     
  13. May 19, 2005 #12

    arildno

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    Also a very good choice, and perhaps the simplest so far.
     
  14. May 19, 2005 #13

    GCT

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    can you clarify? with simplification it seems to lead us right back to trig substitution, or rather a similar form. Perhaps I'm missing something here.
     
  15. May 19, 2005 #14

    arildno

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    Sure:
    [tex]x=\frac{1}{t}\to{dx}=-\frac{dt}{t^{2}}[/tex]
    Thus, we have:
    [tex]\int\frac{dx}{x^{2}\sqrt{1+x^{2}}}=-\int\frac{\frac{dt}{t^{2}}}{\frac{1}{t^{2}}\sqrt{1+\frac{1}{t^{2}}}}=-\int\frac{tdt}{\sqrt{1+t^{2}}}=-\sqrt{1+t^{2}}+C=-\frac{\sqrt{x^{2}+1}}{x}+C[/tex]
    which is at least as simple as the trig. or hyp. substitutions.
     
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