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x ln(x+2) - x + C .... maybe?
let u = x + 2
The way i am looking at it would end up as (1/x+2)
zbay, thats derivative I think ;)
let u=x+2. so du=dx. therefore we have int(lnu)=u(lnu)-u
<---- I am a big dummy 8(
when the sign say push, you pull right ;)
You're on to me already
myin, thanx; thats what I was thinking...
looks fine. I would have just a u substitution. let u=x+2 so du=dx Since u=x+2 then x=u-2 which if we multiply by 6 we have 6x=6(u-2). so we have int(6(u-2)/u,u)=6int(1-2/u,u)=6int(1,u)-6int(2/u,u)=6u-12lnu+C =6(x+2)-12(x+2)+C
oops thats suppose to read 6(x+2)-12ln(x+2)+C
.... i still need a looooottt of practice with integrals ;) thanx
you're super smart amistre. where do you go to college?
down here in florida; at the local community college
finally get to take calc1 and stats during the summer.... if the class dont get canceled that is
mathematics is a huge. i still don't know alot of it, and I have a masters
:) I am working on my masters in math.... i figure at least 4 more years ..
I want to get my phd in applied science. I am going to do a research in cryptography with someone in the computer science department who knows less math than me, so it should be interesting since I don't know alot of computer science. I'm mainly interested in the theory of computer science.
thats sounds fun :) I had started out thinking of getting a computer degree since I write my own programs... but math is my best subject since I was a kid :)