Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

How would you Integrate this?

  1. Nov 29, 2006 #1
    integral of dx/x^(2)-6x +82

    I know from completing the square you get the integral of dx/73 + (x-3)^(2). Then what?
     
  2. jcsd
  3. Nov 29, 2006 #2
    Do the substitution x-3 = u and see if it turns into something a little more familiar.
     
  4. Nov 29, 2006 #3
    I just wanna know what you guys think as there is a discrepancy between my professors answer and my answer.
     
  5. Nov 29, 2006 #4
    What's the discrepancy?
     
  6. Nov 29, 2006 #5
    my professor says that the final answer involves tan^(-1). I say that its (1/73)(u^(-2)du) = (1/73)(-(x-3)^(-1).
     
  7. Nov 29, 2006 #6
    Your professor is right. Check your work again, the denominator is u^2 + 73
     
  8. Nov 29, 2006 #7
    [tex]\int \frac{du}{u^{2}+A} = \int \frac{\sqrt{A}dy}{Ay^{2}+A} = \frac{1}{\sqrt{A}}\int \frac{dy}{y^{2}+1}[/tex]

    Combined with what others have said and what you should have in your notes you will be able to see the professor is right.
     
  9. Nov 29, 2006 #8
    [tex] \int \frac{dx}{x^{2}-6x+82} \; [/tex]


    [tex] \int \frac{dx}{(x-3)^{2} + 73} [/tex]


    Use the fact that [tex]\int \frac{dx}{x^{2}+a^{2}} = \frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right) + C [/tex]

    So going back to the integral:

    [tex] \int \frac{dx}{(x-3)^{2} + 73} [/tex], and [tex] u = x-3, \; du = dx [/tex]

    [tex] a^{2} = 73\Rightarrow a = \sqrt{73} [/tex]

    Thus [tex] \int \frac{dx}{(x-3)^{2} + 73} = \frac{1}{\sqrt{73}}\tan^{-1}\left(\frac{x-3}{\sqrt{73}}\right) + C [/tex]
     
    Last edited: Nov 29, 2006
  10. Nov 29, 2006 #9
    thanks people
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: How would you Integrate this?
Loading...