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!

Indefinite Integral - By parts works right?

  1. Feb 4, 2010 #1
    Nevermind, it's late and I realized why it doesn't work because I forgot to take into consideration that the denominator is (1/polynomial)

    Anyone care to explain to me how to do it the proper way?



    1. Question 1

    [tex]\int[/tex] (x+2)/(x²+x+1) dx

    The only reason I ask is because my teacher showed us another way which basically involves breaking the numerator up, and completing the square in the bottom... which I really don't care to learn. The answer DOES look nicer, but I just want to make sure my way is correct also.

    3. The attempt at a solution

    uv - [tex]\int[/tex]vdu

    u = x+2
    du = 1

    dv = x²+x+1
    v = (1/3)x^3 + (1/2)x^2 + x

    (x+2)[(1/3)x^3 + (1/2)x^2 + x] - [tex]\int[/tex] [ (1/3)x^3 + (1/2)x^2 + x ] (1)

    (x+2)[(1/3)x^3 + (1/2)x^2 + x] - [(1/4)x^4 -(1/6)x^3 + (1/2)x^2] + C

    As I said... Not really pretty by any means, but isn't that technically also correct?[/s]

     
  2. jcsd
  3. Feb 4, 2010 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi RoganSarine! :smile:

    (x+2)/(x²+x+1)

    = (x+2)/((x + 1/2)² + 3/4)

    = (x + 1/2)/((x + 1/2)² + 3/4) + (3/2)/((x + 1/2)² + 3/4) …

    now integrate (you'll need a different method for each part :wink:)
     
  4. Feb 4, 2010 #3
    I can see where to get most of that equation but that last part:

    (x+2)/(x²+x+1)

    (x+1/2+3/2)/(x²+1/2+1/4-1/4)+1)
    (x+1/2+3/2)/((x+1/2)²+1-1/4)
    (x+1/2+3/2)/((x+1/2)²+3/4)


    But...Oh nevermind.

    Wow, ha ha... I've never had to break up an equation like that before (like, breaking up the numerator also, seriously?). I'll give it a shot.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Indefinite Integral - By parts works right?
  1. Indefinite Integral (Replies: 2)

Loading...