#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...

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

#3

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.

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 ****

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?

Can you do it by parts?

#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.

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

#10

E=mc2 lulz0r!

#11

i think i got it now

thanks everyone (except the guy who got so pissed over nothing^^)

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.

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

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

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.