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Calc II: Integration problem

  1. Jan 12, 2010 #1
    1. The problem statement, all variables and given/known data

    Evaluate the indefinite integral:

    2. Relevant equations

    [tex]\int\frac{6}{x\sqrt{25x^2-1}} dx[/tex]

    3. The attempt at a solution

    I've tried a number of different substitutions, but this last one gets me the closest, I think.

    [tex]\int\frac{6}{x\sqrt{25x^2-1}} dx = \frac{6}{5}\int\frac{1}{x\sqrt{x^2-\frac{1}{25}}} dx = \frac{6}{5}\int\frac{x}{x^2\sqrt{x^2-\frac{1}{25}}} dx[/tex]

    let [tex]u=\sqrt{x^2-\frac{1}{25}[/tex]


    [tex]du=\frac{x dx}{\sqrt{x^2-\frac{1}{25}}}}}[/tex]

    This is where I want to substitute in, but it doesn't go. [tex]\frac{x}{\sqrt{x^2-\frac{1}{25}}}}}}dx[/tex] fits in, except for the x^2 in the denominator which doesn't work with my substitution. I don't see where to go from here, or what I should have done differently.

  2. jcsd
  3. Jan 12, 2010 #2


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    Homework Helper

    Try a trig substitution x=cosecθ
  4. Jan 12, 2010 #3
    You're almost done: after substituting x^2=u^2+1/25 in the last integral you obtained, just bring 5 back into the integrand and then you need another substitution, basically, think of u=\sqrt(5)*cosh(x). Then you'll be left with a simple integral which is neccessarily required to be given a form like e^t/(1+e^2t) dt. From this point, you need one more substitution and then you are done by going back to x as you begin from this last one, pass t and u to definitely hit x again.

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