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!

Minus sign before the 2nd integral

  1. Dec 6, 2007 #1
    [Solved] Where did I go wrong?

    [tex]\int \frac{(x-2)}{x(1+x^2)}dx[/tex]

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

    I choose [tex] u = arctan(x) [/tex] so [tex] du = \frac{1}{1+x^2} [/tex]

    and [tex] \int \frac{x}{x+x^3} = \int \frac{1}{1+x^2} [/tex]

    1)

    [tex] \int \frac{1}{1+x^2}dx = arctan(x) [/tex]

    2)

    [tex] \int \frac{2}{x+x^3} = \ln (\frac{x^2}{x^2+1}) [/tex]

    So [tex] \int \frac{(x-2)}{x(1+x^2)}dx = arctan(x) + \ln (\frac{x^2}{x^2+1}) [/tex]


    Maybe I'm doing something terribly wrong but maybe not. The answer is according to Texas Instrument calculator:

    [tex] \ln (\frac{x^2+1}{x^2}) + arctan(x) [/tex]
     
    Last edited: Dec 6, 2007
  2. jcsd
  3. Dec 6, 2007 #2

    Shooting Star

    User Avatar
    Homework Helper

    There's a minus sign before the 2nd integral.
     
  4. Dec 6, 2007 #3
    Ok thanks for that correction. But what about the 2nd integral? How is it done ? The denominator and numerator are switched if you look at my answer and the TI answer.
     
  5. Dec 6, 2007 #4

    Shooting Star

    User Avatar
    Homework Helper

    The minus sign makes your answer correct because -log x = log 1/x.

    How to get the integral -- well, give me some time.
     
  6. Dec 6, 2007 #5

    Shooting Star

    User Avatar
    Homework Helper

    2dx/x(1+x^2) = 2xdx/x^2(1+x^2).

    Put v=x^2 => dv=2xdx. The integral becomes dv/v(1+v). That's simple, I hope.
     
  7. Dec 6, 2007 #6
    okey. I tried this and although it looks very logical to me this is not correct??

    I got int dv/2v = ln(x^2)/2

    any ideas?
     
  8. Dec 6, 2007 #7

    Shooting Star

    User Avatar
    Homework Helper

    Incorrect. Try the substitution I had shown in my last post.

    dv/v(1+v) = [(1+v)-v]dv/v(1+v). Can you complete it?
     
  9. Dec 6, 2007 #8
    doh. I see it now. But I still get [tex] \ln (\frac{x^2}{x^2+1}) [/tex]
     
  10. Dec 6, 2007 #9

    Shooting Star

    User Avatar
    Homework Helper

    There's a minus sign before this one.
     
  11. Dec 6, 2007 #10
    What does that change? I thought that it would be the same result but a negative one.
     
  12. Dec 6, 2007 #11
    OK! I get it! It changes everything!

    Thank you very much.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Minus sign before the 2nd integral
Loading...