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Integral of a function

  1. Feb 26, 2006 #1
    [tex]\int \frac{dx}{\sqrt{1/x + C}}[/tex] where C is a constant. Any ideas?
     
  2. jcsd
  3. Feb 26, 2006 #2

    arildno

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    Rewrite your integrand as:
    [tex]u=\sqrt{\frac{x}{1+Cx}}[/tex]
    Try and integrate with respect to u; that usually works in cases like this.
     
  4. Feb 27, 2006 #3

    HallsofIvy

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    Is that [itex]\frac{1}{x}+ C[/itex] or [itex]\frac{1}{x+C}[/itex]?
     
  5. Feb 27, 2006 #4

    benorin

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    The former, viz. [tex]\int \frac{1}{\sqrt{\frac{1}{x}+C}} dx = \int\sqrt{\frac{x}{1+Cx}} dx,[/tex] is nasty looking and the later is quite simple, of chourse it is just [tex]\int\sqrt{x+C}dx = \frac{2}{3}(x+C)^{\frac{3}{2}}+C_1[/tex]
     
  6. Feb 27, 2006 #5
    Maybe a little longer approach,
    But this is the way I’d approach this in a cal 2 class
    Please excuse the notation.

    First rewrite integral sqrt(X)/sqrt(1+CX) dx

    Now let 1/x =u^2
    dx=-2/(1/u^3)du
    also let a^2 = c to make notation a little easier to work with.

    This will yield an integral that look likes within constants
    1/(u^3 sqrt(u^2+a^2) ) du

    Now use a standard trig substitution let tan(angle)= a^2 u^2
    the top numerator is just sec^2(angle) *d (angle)
    the bottom portion is tan(angle)^3 * sec(angle)

    Now simplify
    We are integrating to within a constants sec(angle)/tan^2(angle) * d(angle)

    Still have some simplification to do using the definition of tan = sin/cos
    And sec=1/cos.

    The form reduces to cos(angle) d(angle) / sin^2 (angle)
    The integral is of this form is 1/sin^3(angle) and now all you need to do is all the back substitution.

    And you can now begine to see the (2/3)(X+C)^(3/2) form reported by Benorin

    If I was teaching cal 2 ( I’m doing cal 1) this semester I’d consider this a great take home problem for credit.

    A rewrite would be nice if someone has a program which writes in a better math notation.

    Again this is a great problem!
     
  7. Feb 27, 2006 #6

    arildno

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    To give a few hints on my approach, we start with:
    [tex]u=\sqrt{\frac{x}{1+Cx}}\to{u}^{2}=\frac{x}{1+Cx}\to{x}=\frac{u^{2}}{1-Cu^{2}}[/tex]

    We therefore have:
    [tex]\frac{dx}{du}=\frac{2u(1-Cu^{2})+2Cu^{3}}{(1-Cu^{2})^{2}}=\frac{2u}{(1-Cu^{2})^{2}}[/tex]
    by which it follows:
    [tex]\int\sqrt{\frac{x}{1+Cx}}dx=\int\frac{2u^{2}}{(1-Cu^{2})^{2}}du[/tex]
    The latter integral is easily calculated.
     
    Last edited: Feb 27, 2006
  8. Feb 27, 2006 #7
    Easily??
    Well, the previous ones seem to be wrong, I got this monster from mathematica:

    [tex]\frac{cx+1}{c \sqrt{c + \frac{1}{x}}} - \frac{\sqrt{cx+1} asinh {\sqrt{cx}} }{c^{3/2} \sqrt{c + \frac{1}{x}} \sqrt {x}}[/tex]

    BTW, how did you guys derieved [tex]\int\sqrt{x+C}dx[/tex] from [tex]\int\sqrt{\frac{x}{1+Cx}} dx[/tex]?
     
    Last edited: Feb 27, 2006
  9. Mar 1, 2006 #8

    VietDao29

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    No, we did not say that:
    [tex]\int \sqrt{x + C} dx = \int \sqrt{\frac{x}{1 + Cx}} dx[/tex]
    What benorin says is just that, if the original integral is:
    [tex]\int \frac{dx}{\sqrt{\frac{1}{x + C}}}[/tex], then it can be rearranged to give:
    [tex]\int \frac{dx}{\sqrt{\frac{1}{x + C}}} = \int \sqrt{x + C} dx[/tex].
    :)
     
  10. Mar 1, 2006 #9
    Whoops! :)
    Hmm, well then, what if it was [tex]\int\sqrt{\frac{x}{1+Cx}} dx[/tex]? If what mathematica says is true, then it's gonna be a BIG headache to solve. But I want to see what this headache is!
     
  11. Mar 1, 2006 #10

    HallsofIvy

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    Partial fractions is not a big headache. Tedious perhaps.
     
  12. Mar 2, 2006 #11

    arildno

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    Hint:
    Use integration by parts first:
    [tex]\int\frac{2u^{2}}{(1-Cu^{2})^{2}}du=\frac{u}{C(1-Cu^{2})}-\frac{1}{C}\int\frac{du}{1-Cu^{2}}[/tex]
    Use partial fractions on the latter integral, and you're done
     
    Last edited: Mar 2, 2006
  13. Mar 8, 2006 #12
    How can you write 1 + Cx when the original is 1/x + C? Pardon my ingnorance on this subject, but we just started these last week.
     
  14. Mar 8, 2006 #13

    arildno

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    Very good question, misskitty!
    Remember that the original integrand is [itex]\frac{1}{\sqrt{\frac{1}{x}+C}}[/itex]
    Now, we may rewrite this as:
    [tex]\frac{1}{\sqrt{\frac{1}{x}+C}}=\sqrt{\frac{1}{\frac{1}{x}+C}}=\sqrt{\frac{x}{x}*\frac{1}{\frac{1}{x}+C}}=\sqrt{\frac{x}{Cx+1}}[/tex]
     
  15. Mar 8, 2006 #14
    OH!!! So you CAN do that because all you've done is multiplied by a name for one inside the radical! That would have made one of my problems so much easier. Thanks for explaining. :biggrin:

    ~Kitty
     
  16. Mar 8, 2006 #15
    Also what's the purpose of changing the variable from x to u? What's the difference between integrating with respect to x or respect to u? Its still integrating with respect to a variable.

    ~Kitty
     
  17. Mar 8, 2006 #16

    arildno

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    Note that the u-substitution transforms your integral from an ugly expression involving a square root into a rational function (i.e, whose numerator and denominator are polynomials).
    Since we know how to integrate rational functions, we've just made our original problem easily solvable..
     
  18. Mar 8, 2006 #17
    So by replacing it with u then you can just deal with the numbers without the square root and then once everything is worked out then you can take the square root. Do I have this correct?

    ~Kitty
     
  19. Mar 8, 2006 #18

    arildno

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    Yes, having found your anti-derivative in u, just replace every u through [itex]u=\sqrt{\frac{x}{Cx+1}}[/itex], and you're done.
     
  20. Mar 8, 2006 #19
    I get it now! Yay. Thanks for the patience to explain it. :biggrin:

    ~Kitty
     
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