Integral (2x+4)/(x^2+2x+3) w.r.t. x

[dmission]

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

 

[HallsofIvy]

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]

\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]

thanks for the reply.

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

 

[rock.freak667]

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

 

[dmission]

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

 

[rock.freak667]

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 ?

 

[dmission]

bump, anyone?