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Intergration of to functions:

  1. Apr 27, 2007 #1
    Hello, I'm having difficulty intergrating the following two functions:

    1) (6x-x^2)^(-1/2)

    I'm fairly sure that you need to intergrate by parts for this function, but I can't figure out how to do it

    2) e/(e^(2x)-1)

    This one I thought could do with inverse trig functions but I'm drawing a blank there also

    Can someone please help me???
  2. jcsd
  3. Apr 28, 2007 #2

    Gib Z

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    1) [tex]\int \frac{1}{\sqrt{6x-x^2}} dx[/tex]

    Whenever I see something like that, I take out as much of the x's from the square root as possible. So lets take out an x^2 :)

    [tex]\int \frac{1}{\sqrt{6x-x^2}} dx = \int \frac{1}{\sqrt{x^2}\sqrt{\frac{6}{x} -1}}[/tex]

    Now thats starting to look like something a trig substitution can help :)

    I seem to remember [itex]\sec^2 u -1 = \tan^2 u[/itex] :)

    So to make it very convenient, the 6/x would be nice to be sec^2 u.

    So let x= 6 cos^2 u. Thats a nice substitution :)

    2)[tex]\int \frac{e^x}{(e^x)^2-1} dx[/tex]

    u=e^x, du = e^x dx

    So it becomes [tex]\int \frac{1}{u^2-1} du[/tex]

    To finish that off, realise the bottom factors by difference of two squares into (u+1)(u-1) and use partial fractions.
    Last edited: Apr 28, 2007
  4. Apr 28, 2007 #3
    Thank you so much for your help, I really appreciate it
  5. Apr 28, 2007 #4


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    A shorter version is:

    [tex]\int \frac{dx}{\sqrt{6x - x ^ 2}} = \int \frac{dx}{\sqrt{- \left( -6x + x ^ 2 \right)}} = \int \frac{dx}{\sqrt{9 - \left( x ^ 2 - 6x + 9 \right)}} = \int \frac{dx}{\sqrt{9 - \left( x - 3 \right) ^ 2}} = ...[/tex]


    Usually, when you encounter the form:
    [tex]\int \frac{dx}{\sqrt{Ax ^ 2 + Bx + C}}[/tex]
    We divide it into 2 cases:
    The first case is A > 0, we'll try to arrange the integral above to the form:
    [tex]\int \frac{dx}{\sqrt{Ax ^ 2 + Bx + C}} = \int \frac{dx}{(mx + n) ^ 2 + \alpha ^ 2}[/tex]
    Then use the substitution:
    [tex]mx + n = \alpha \tan t[/tex]

    The second case is A < 0, we'll re-arrange it to:
    [tex]\int \frac{dx}{\sqrt{Ax ^ 2 + Bx + C}} = \int \frac{dx}{\alpha ^ 2 - (mx + n) ^ 2}[/tex]
    Then use the substitution:
    [tex]mx + n = \alpha sin t[/tex]


    Is it clear? Can you complete the problem now? :)
    Your problem is in the second case. :smile:
  6. Apr 30, 2007 #5
    Well I'm having difficulty with the parts problem now because you must make the denominator equal to zero to solve for A and B to solve the equation, but e^x + 1 never equals zero so I'm a little confused how to solve it???
  7. May 1, 2007 #6

    Gib Z

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    Does that help?
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