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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


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    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


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    [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


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    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


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    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?
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