Integral question

1. Nov 25, 2008

dmission

how:
integral: (2x+4)/(x^2+2x+3) with respect to x, of course

2. Nov 26, 2008

HallsofIvy

Complete the square in the denominator. They try something! We are not going to do your homework for you.

3. Nov 26, 2008

dmission

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. Nov 26, 2008

lucidicblur

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. Nov 27, 2008

dmission

what do you mean by take out the constants of integration?

6. Nov 29, 2008

dmission

bump, really not sure what to do :S

7. Nov 29, 2008

rock.freak667

$$\int \frac{2x+4}{x^2+2x+3} dx = \int \frac{2x+4}{(x+1)^2+2} dx$$

$$= \int \frac{2x}{(x+1)^2+2}dx + \int \frac{4}{(x+1)^2+2}}dx$$

Now just recall that:

$$\int \frac{1}{X^2+A^2} dx = \frac{1}{A}tan^{-1}(\frac{X}{A})+k$$

8. Nov 29, 2008

dmission

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

9. Nov 29, 2008

rock.freak667

Try a substitution of u=x+1

10. Nov 29, 2008

dmission

yeah, but doesn't that still leave an X in the numerator?

11. Nov 29, 2008

rock.freak667

u=x+1 => x=u-1

12. Nov 29, 2008

dmission

ugh, never knew about that... can you please work out that portion for me? really not sure what to do...

13. Nov 29, 2008

dmission

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. Nov 30, 2008

dmission

bump, anyone?