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Calculate an integral using euler substituition.

  1. Jun 9, 2007 #1
    i was asked to calculate the integral:
    [tex]\int\frac{dx}{x+\sqrt{x^2-x+1}}[/tex] by using euler substituition (i.e, finding a line which intersects sqrt(x^2-x+1) through one point and then the equation of the line will be y-y0=t(x-x0) where (x0,y0) is one point of intersection, and then substituing x for a rational function of t.
    i did so in this particular example but i got stuck.
    so i tried to recheck it through mathworld's integrator, but it states there isn't such formula for this integral, is this correct?
    obviously if it's integrator, so it must be. (-:
     
  2. jcsd
  3. Jun 9, 2007 #2

    VietDao29

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    Homework Helper

    Well, I'd do it this way, let t, such that:
    [tex]\sqrt{x ^ 2 - x + 1} = t - x \quad (1)[/tex] or stated differently [tex]t = x + \sqrt{x ^ 2 - x + 1}[/tex]

    Now, we'll square both sides of (1):
    [tex]\Rightarrow x ^ 2 - x + 1 = t ^ 2 - 2xt + x ^ 2[/tex]
    [tex]\Rightarrow - x + 1 = t ^ 2 - 2xt[/tex]

    Isolate x's to one side, and we'll try to express x in terms of t:
    [tex]\Rightarrow - x + 2xt = t ^ 2 - 1[/tex]

    [tex]\Rightarrow x (2t - 1) = t ^ 2 - 1[/tex]

    [tex]\Rightarrow x = \frac{t ^ 2 - 1}{2t - 1} \quad (2)[/tex]

    Take the differential of both sides yields:

    [tex]\Rightarrow dx = \frac{2 t ^ 2 -2t + 2}{(2t - 1) ^ 2} dt = 2 \frac{t ^ 2 - t + 1}{(2t - 1) ^ 2} dt \quad (3)[/tex]

    Now substitute (1), (2), and (3) to your original integral, we'll have:
    [tex]\int \frac{dx}{x + \sqrt{x ^ 2 - x + 1}} = 2 \int \frac{\frac{t ^ 2 - t + 1}{(2t - 1) ^ 2} dt}{t} = 2 \int \frac{t ^ 2 - t + 1}{(2t - 1) ^ 2 t} dt[/tex]
    It looks much better than its original form, right?
    Can you go from here? :)
     
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