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!

Homework Help: Integral question

  1. Nov 25, 2008 #1
    integral: (2x+4)/(x^2+2x+3) with respect to x, of course
  2. jcsd
  3. Nov 26, 2008 #2


    User Avatar
    Science Advisor

    Complete the square in the denominator. They try something! We are not going to do your homework for you.
  4. Nov 26, 2008 #3
    I tried -- best I could do in the denominator was (x+1)^2 +2, still not sure where to go from there though.

    any help would be appreciated
  5. Nov 26, 2008 #4
    your completing the square part is fine. after this you will need to
    -spilt up the fraction
    -take out the constants of integration
    -do a u substitution for the 2x part of numerator
    -use arctan for the 4 part of the numerator

    you will notice the the part you completed the square looks almost like the arctan formula except you will need to change 2 into (sqrt2)2 that way everything in the denominator is squared.
  6. Nov 27, 2008 #5
    what do you mean by take out the constants of integration?
  7. Nov 29, 2008 #6
    bump, really not sure what to do :S
  8. Nov 29, 2008 #7


    User Avatar
    Homework Helper

    [tex]\int \frac{2x+4}{x^2+2x+3} dx = \int \frac{2x+4}{(x+1)^2+2} dx[/tex]

    [tex]= \int \frac{2x}{(x+1)^2+2}dx + \int \frac{4}{(x+1)^2+2}}dx[/tex]

    Now just recall that:

    [tex]\int \frac{1}{X^2+A^2} dx = \frac{1}{A}tan^{-1}(\frac{X}{A})+k[/tex]
  9. Nov 29, 2008 #8
    thanks for the reply.

    I get how to take the second one, but what about the first? (2x/((x+1)^2+2))
  10. Nov 29, 2008 #9


    User Avatar
    Homework Helper

    Try a substitution of u=x+1
  11. Nov 29, 2008 #10
    yeah, but doesn't that still leave an X in the numerator?
  12. Nov 29, 2008 #11


    User Avatar
    Homework Helper

    u=x+1 => x=u-1 :smile:
  13. Nov 29, 2008 #12
    ugh, never knew about that... can you please work out that portion for me? really not sure what to do...
  14. Nov 29, 2008 #13
    So, tried, got:
    2 * integral: (u-1)/(u^2+2), split up again,
    and ultimately got:

    ln((x+1)^2+2) - 2/sqrt(2)*arctan((x+1)/sqrt(2))), but apparently i'm wrong... help ?
  15. Nov 30, 2008 #14
    bump, anyone?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook