1. Limited time only! Sign up for a free 30min personal 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!

Integral and Arctangent

  1. Oct 1, 2014 #1
    What did I do wrong evaluating this integral: ∫ [4 / (4x^2 + 4x + 65) dx ?



    Here were my steps:
    1) = 4 ∫ [1 / (4x^2 + 4x + 65)] dx
    2) = 4 ∫ [1 / (4x^2 + 4x + 1 + 64)] dx
    3) = 4 ∫ [1 / ((2x+1)^2 + 64)] dx ;

    Then I applied the formula: ∫ [1 / √(a^2 + u^2)] dx = (1/a)arctan(u/a) + C;

    4) = (4)(1/8)arctan [(2x+1) / (√64)] + C
    5) = (1/2)arctan [(2x+1) / 8] + C

    My answer: (1/2)arctan [(2x+1) / 8] + C



    The correct answer: (1/4)arctan[(2x+1) / 8] + C
     
  2. jcsd
  3. Oct 1, 2014 #2

    Mark44

    Staff: Mentor

    Your answer is too large by a factor of 2. I suspect that when you made the substitution from x to u, you forgot to change from dx to du correctly. If u = 2x + 1, then du = 2*dx.

    BTW, when you do an integration problem, it is a good idea to check your work by differentiating your answer. You should end up with the same integrand you started with.
     
  4. Oct 2, 2014 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    There is no such formula. Did you mean "[itex]\int \frac{1}{a^2+ u^2} du= arctan(u/a)+ C[/itex]"?

     
    Last edited: Oct 2, 2014
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted