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Integral question

  • Thread starter dmission
  • Start date
10
0
how:
integral: (2x+4)/(x^2+2x+3) with respect to x, of course
 

HallsofIvy

Science Advisor
Homework Helper
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Complete the square in the denominator. They try something! We are not going to do your homework for you.
 
10
0
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
 
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.
 
10
0
what do you mean by take out the constants of integration?
 
10
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bump, really not sure what to do :S
 

rock.freak667

Homework Helper
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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]
 
10
0
thanks for the reply.

I get how to take the second one, but what about the first? (2x/((x+1)^2+2))
 

rock.freak667

Homework Helper
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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?
 
10
0
ugh, never knew about that... can you please work out that portion for me? really not sure what to do...
 
10
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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 ?
 
10
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bump, anyone?
 

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