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How to deal w/ inverse trig function within ∫(x)/((x^2-4)(√(x^4 -8)))?

  1. Sep 4, 2012 #1
    Evaluate the integral [tex] \int \frac{x}{(x^2 - 4) \sqrt{x^4 - 8x^2}} \, dx [/tex]

    by making the substitution [itex]u = x^{2}[/itex]


    2. Relevant equations





    3. The attempt at a solution


    [itex]u = x^{2} - 4[/itex] so [itex]\frac{du}{2}= xdx[/itex]

    [tex] \int \frac{1}{(x^2 - 4) \sqrt{x^4 - 8x^2}} \, xdx [/tex]


    [tex]\frac{1}{2} \int \frac{1}{(u) \sqrt{x^4 - 8x^2}} \, du[/tex]


    [tex]\frac{1}{2} \int \frac{1}{(u)} \frac{1}{\sqrt{x^4 - 8x^2}} \, du[/tex]

    [tex]\frac{1}{2} \int \frac{1}{(u)} \frac{1}{|x|\sqrt{x^2 - 8}} \, du[/tex]


    I haven't yet committed all of the inverse trig functions to memory but I have a table that shows dy/d[itex]sec^{-1}(x) = \frac{1}{|x| \sqrt{x^2 - 1}}[/itex]

    However, I do not know how to handle [itex]\frac{1}{|x| \sqrt{x^2 - 8}}[/itex]

    What does that equal and why? I mean what is the relationship between it and whatever it's antiderivative may be?
     
    Last edited: Sep 4, 2012
  2. jcsd
  3. Sep 4, 2012 #2

    vela

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    Try doing it in two steps. Use the hint and make the first substitution u=x2. Rewrite everything in terms of u. Then complete the square inside the radical and go from there.
     
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