#1
i guess this attempt's pointless but anyway :

someone got an idea what the integral of (2e^x) / (2 + e^x)^2 is?
out teacher told us to use substitution but i just can't seem to solve this one...
#2
yes im good maths took my a levels in year 9, and if u hadnt of said maths geek i would of helped you, u can be inteligent and not be a noobing geek, douche bag
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#3
Quote by GuitarManDan15
yes im good maths took my a levels in year 9, and if u hadnt of said maths geek i would of helped you, u can be inteligent and not be a noobing geek, douche bag


Lol, be proud to be a geek instead of feeling insulted!
#4
let

u=2 + Exp(x)
du=Exp(x)dx

inserting gives

Integral{ (2 / u^2)*du } = 2 * Integral{ du/(u^2) }

Now solve that.
#5
wooohooo theres somebody who is easiyl offended...
actually i'm good at informatics physics and math (even if i cant just get this integral right, all A's this year (13th grade)) and people sometimes call me geek but i dont give a ****
#6
Man, that's hard.
Can you do it by parts?
I hope it doesn't seem, like I'm young, foolish, and green.
Let me in for a minute, you're not my life but I want you in it


O Dayya, te echaré de menos, siempre

Y siempre
Y para siempre
#7
thx @ logic Smogic i already did it that way in the end i get : - 1 / (2 + e^x) but that unfortunately doesnt seem to be right
#8
Realize that using my substitution, you just need to find the integral of 1/u^2, and then insert u=2*Exp(x).

EDIT: ^ Oh? ...hmm

EDIT2: (down below).. yeah, the Integrator says log.
Last edited by Logic Smogic at Feb 13, 2008,
#9
ah goddammit i found a mistake : i put a constant in front of the integral sign and forgot to calculate it in in the end... damn
#11
i think i got it now
thanks everyone (except the guy who got so pissed over nothing^^)
#12
No problem. I almost never have to do integrals these days (Dirac does'em for me!).. so it's good practice to do stuff like this.
#13
That might take a while.

If you understand the chain rule, use it on the top and bottom of the fraction.
y=[f(x)]^n then dy/dx= n[f(x)]^(n-1) [f'(x)]

Then differentiate the whole thing using quotient rule.

Quotient rule: y= f/g then dy/dx= [g(x)f'(x)-f(x)g'(x])/[g(x)]^2

I'm going to assume that you know how to differentiate e.
Last edited by SmarterChild at Feb 13, 2008,
#14
Quote by GuitarManDan15
yes im good maths took my a levels in year 9, and if u hadnt of said maths geek i would of helped you, u can be inteligent and not be a noobing geek, douche bag


Maybe you're good at math but where is the grammar?

And I don't even know what an integral is... I thought I was pretty good at math too.. I'm in 9th grade. Am I supposed to know what that is?...
#15
Quote by Jonjy2
Maybe you're good at math but where is the grammar?

And I don't even know what an integral is... I thought I was pretty good at math too.. I'm in 9th grade. Am I supposed to know what that is?...


No, I didn't take calculus until I was a senior in High School, and now I'm a physicist. So no, I wouldn't worry about anything.