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Homework Help: How can I find te integral of root(x*x+x+1)

  1. Jan 2, 2005 #1
    How can I find te integral of root(x*x+x+1)Thank you
  2. jcsd
  3. Jan 2, 2005 #2


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    x2+ x+ 1= x2+ x+ 1/4+ 3/4= (x+ 1/2)2+ 3/4.

    Let u= x+ 1/2.
  4. Jan 2, 2005 #3


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    You need to put this integral in the form of

    [tex] \int \sqrt{a^2 + u^2} du [/tex]

    This can be achieved by completing the square.
  5. Jan 2, 2005 #4


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    I'm assuming you're talking about this "animal".
    [tex] I=\int \sqrt{x^{2}+x+1} dx =\int \sqrt{(x+\frac{1}{2})^{2}+\frac{3}{4}} dx [/tex] (1)

    Like Halls said,make the first substitution [itex] x+\frac{1}{2}\rightarrow u [/itex] (2)
    and u'll get
    [tex] I=\int \sqrt{u^{2}+\frac{3}{4}} du [/tex] (3)
    Write this integral as
    [tex] I=\int \sqrt{\frac{3}{4}} \sqrt{(\sqrt{\frac{4}{3}} u)^{2} +1} du [/tex] (4)
    and make the substitution
    [tex] \sqrt{\frac{4}{3}} u\rightarrow \sinh v [/tex] (5)
    ,under which:
    [tex] du\rightarrow \sqrt{\frac{3}{4}} \cosh v dv[/tex] (6)
    ,so the integral becomes:
    [tex] I=\frac{3}{4}\int \cosh^{2}v dv [/tex] (7)
    ,where i have made use of the fundamental formula of hyperbolic trigometry
    [tex] \cosh^{2}v-\sinh^{2}v =1 [/tex] (8)
    Use the formula of the double angle:
    [tex] \cosh^{2}v=\frac{1+\cosh 2v}{2} [/tex](9)
    to solve the integral in terms of 'v'.Return to the initial variable 'x' via the substitutions (5) and (2).

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