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Integration problem help

  1. Apr 2, 2005 #1
    Is there a relatively simple way to solve this integral? Because I feel like I should know how to solve it, but I can't think of any way to do it.

    [tex]\lim_{n\rightarrow\infty}\int_{1}^{n} \frac{x\,dx}{x^4+1}[/tex]
  2. jcsd
  3. Apr 2, 2005 #2
    Factor the denominator of the integrand:

    [tex]x^4 + 1 = (x^2 - \sqrt{2}x + 1)(x^2 + \sqrt{2}x + 1)[/tex]

    then use partial fractions to simplify and integrate. Don't worry about the limit until after that... your expression is really the same as

    [tex] \int_1^\infty \frac{xdx}{x^4+1}[/tex]
  4. Apr 2, 2005 #3
    Or on the other hand you could just sub [itex] u = x^2[/itex] without doing anything~

    I'm silly!
  5. Apr 2, 2005 #4
    This is the arctangent rule.

    Let u = x^2
    du = 2xdx
    du / 2 = xdx

    [tex]\int_{1}^{n}\frac{xdx}{x^4+1} = \frac{1}{2}\int \frac{du}{u^2+1}[/tex]

    [tex] = \frac{\arctan{u}}{2} = \frac{\arctan{x^2}}{2}[/tex]

    Now apply the bounds and take the limit.
  6. Apr 2, 2005 #5
    [tex]\lim_{n \to \infty} \int_1^n \frac{xdx}{x^4+1}[/tex]
    and you guys got this is equal to:
    [tex]\lim_{n \to \infty}\frac{arctan(n^2)}{2}-\frac{arctan(1)}{2}=\lim_{n \to \infty}\frac{arctan(n^2)}{2}-0.3926990815=0.785398165 - 0.3926990815=0.3926990835[/tex]

    i think thats it. :rofl: i dont have a book or anything, so i dont really know. hopefully that helps.
  7. Apr 2, 2005 #6


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    You needn't have put in decimal form (BTW,both [itex] \frac{\pi}{4} [/itex] and [itex] \frac{\pi}{8} [/itex] have an infinite # of decimals)...

    So the integral is

    [tex] \int_{1}^{+\infty} \frac{x}{x^{4}+1} \ dx =\frac{\pi}{8} [/tex]

  8. Apr 2, 2005 #7
    heh, thanks for clearing up what i did. all i had was this stupid computers calculator; im at work. im surprised i even got the right answer, haha. :rofl:
  9. Apr 2, 2005 #8


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    Incidentally,your subtraction is wrong :tongue:

  10. Apr 2, 2005 #9
    arggg, its right. look at it!!
    [tex]0.785398165 - 0.3926990815 \approx 0.392699082 \approx \frac{pi}{8}[/tex]
    its legit :grumpy:
  11. Apr 4, 2005 #10


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    Check the last decimals and blame it on the lousy computer...:wink: :tongue:

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