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Indefinate Integrals

  1. Mar 31, 2007 #1
    1. The problem statement, all variables and given/known data
    Hi, I'm having trouble finding the indefinite integral of an equation.

    2. Relevant equations
    The function is (2x-13)/[(x^2-2x+4)^0.5]

    3. The attempt at a solution
    I thought it was a good idea to let u=x^2-2x+4, hence du/dx= 2x-2.
    From that I got [-11/(u^0.5)].du but evaluating this gives me -22(u^0.5), which I don't beleive is right.
    Did I make a silly mistake or is my approach completely wrong? I am also having similar trouble finding the integral of [1+(1/t)]^5/(t^2).

    Any help or advice would be very much appreciated, thank you.
  2. jcsd
  3. Mar 31, 2007 #2
    Your approach is correct, but you need rewrite the integral slightly.


    [tex]\int{\frac{2x-2}{\sqrt{x^2-2x+4}}} + \int{\frac{-11dx}{\sqrt{x^2-2x+4}}}[/tex]

    Now substitute [itex]u=x^2-2x+4[/itex], in the first integral.

    As for the other integral, you can rewrite 1 + (1/t) as (t+1)/t.
    Last edited: Mar 31, 2007
  4. Mar 31, 2007 #3
    for your second q, i would manipulate it to look like (t+1)^5/t^7 and use binominal expansion for the top.
  5. Mar 31, 2007 #4


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    [tex]\int \frac{2x-13}{\sqrt{x^2- 2x+ 4}} dx= \int \frac{2x-2}{\sqrt{x^2- 2x+4}} dx- \int \frac{11}{\sqrt{x^2-2x+ 4}}dx[/tex]
    Since the derivative of x2- 2x+ 4 is 2x- 2, the first can be done with the substitution u= x2-2x+ 4, but the second cannot.

    Rewrite x2- 2x+ 4 as x2-2x+ 1+ 3= (x-1)2+3 (in other words, complete the square), let [itex]v= \sqrt{3}(x-1)[/itex] and then use a trig substitution
  6. Mar 31, 2007 #5
    Or as dexter would say, use a hyperbolic trig substitution. :biggrin:
  7. Mar 31, 2007 #6
    Thank you soo much for all the help guys.
    One question about HallsofIvy's comment. I understand why and how you completed the square. But don't understand why v=sq.rt.3(x-1). Now dv/dx=sq.rt.3. If I substitute this into the denominator I get a function that I don't know what to do with.
    I tried to use standard integrals and basically pulled the 11 in front of the integration sign and was left with 1/[(x-1)^2+(3^0.5)^2]^0.5. Which using standard integral formula I was left with arcsinh[(x-1)/(3^0.5)]+c.
    Again I suspect I am wrong with this approach.
  8. Mar 31, 2007 #7
    (-11)arcsinh[(x-1)/(3^0.5)]+c is correct.
  9. Mar 31, 2007 #8
    Great. Thanks again everyone.
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