Integral question

  • Thread starter dmission
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  • #1
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how:
integral: (2x+4)/(x^2+2x+3) with respect to x, of course
 

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  • #2
HallsofIvy
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Complete the square in the denominator. They try something! We are not going to do your homework for you.
 
  • #3
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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
 
  • #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.
 
  • #5
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what do you mean by take out the constants of integration?
 
  • #6
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bump, really not sure what to do :S
 
  • #7
rock.freak667
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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]
 
  • #8
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thanks for the reply.

I get how to take the second one, but what about the first? (2x/((x+1)^2+2))
 
  • #9
rock.freak667
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thanks for the reply.

I get how to take the second one, but what about the first? (2x/((x+1)^2+2))
Try a substitution of u=x+1
 
  • #10
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yeah, but doesn't that still leave an X in the numerator?
 
  • #11
rock.freak667
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yeah, but doesn't that still leave an X in the numerator?
u=x+1 => x=u-1 :smile:
 
  • #12
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ugh, never knew about that... can you please work out that portion for me? really not sure what to do...
 
  • #13
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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 ?
 
  • #14
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bump, anyone?
 

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