# Integral question

#### dmission

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

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#### HallsofIvy

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

#### 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

#### 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.

#### dmission

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

#### dmission

bump, really not sure what to do :S

#### rock.freak667

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

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

Now just recall that:

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

#### dmission

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

#### rock.freak667

Homework Helper

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

#### dmission

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

#### rock.freak667

Homework Helper
yeah, but doesn't that still leave an X in the numerator?
u=x+1 => x=u-1 #### dmission

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

#### 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 ?

bump, anyone?