1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Staff Emeritus
    Science Advisor

    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


    User Avatar
    Homework Helper

    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


    User Avatar
    Science Advisor
    Homework Helper

    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).

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?