Integration q... im stuck... (1 Viewer)

[AGB]

Hey ppl,

Could someone please help me with this question... integrate the following:

e^-x
-------
1 + e^2x

I just have no idea what to do... everything I try just seems to get nowhere

Thanks

 

[nike33]

= int 1/[e^x(1+e^x)]
= int 1/e^x - int(1/(1+e^x))
which is ezy

 

[turtle_2468]

Hey!
Basically you want to get rid of the top bit, since it's annoying.
So letting u=e^(-x)
du/dx=-e^(-x)
So the integral becomes -Int(u/{1+(1/u)})du
=-Int(u^2/(u+1))du
=-Int(u-1+1/(u+1))du
=-(u^2/2-u+ln(u+1))
=-(u^2/2)+u-ln(u+1)
and sub u=e^(-x) back in.

Shorter is what nike did... although I'd dispute the second one being "easy"..

 

[AGB]

hey,

woops! i made a typo... sorry. the question should read:

integrate:

e^-x
-------
1 + e^2x

thanks

 

[nike33]

notice it equals

1/e^x - e^x/(1+e^2x)

 

[AGB]

would you use partial fractions??

 

[nike33]

not with simple ones like that...just sorta look at it and see waht it equals ie

e-x
-------
1 + e2x

= 1/e^x - e^x/(1+e^2x) which is easy to intergrate

 

[AGB]

Originally posted by nike33
notice it equals

1/e^x - e^x/(1+e^2x)

how did you get that??

EDIT: sorry if it is easy... i just dont understand

thanks for your help as well

 

[nike33]

e^-x = 1/e^x

so e^-x / (1 + e^2x) = 1 / (e^x(1+e^2x))

you can now do partial fractions(have you done this yet? ..but in general when the numerators a low number1,2 etc/ or works out nicely its bettter / easier to just observe (saves time in exams!)

 

[AGB]

Originally posted by nike33
e^-x = 1/e^x

so e^-x / (1 + e^2x) = 1 / (e^x(1+e^2x))

you can now do partial fractions(have you done this yet? ..but in general when the numerators a low number1,2 etc/ or works out nicely its bettter / easier to just observe (saves time in exams!)

i dont get how you have observed it though?? like i know that it works, but how did u get to 1/e^x - e^x/(1+e^2x) just by looking at it??

 

[nike33]

i guess if u practive u just see it ..

 

[Affinity]

he/she might have worked it out on paper and omitted it here or ... it's not that hard to do in the brain for some people

 

[ToO LaZy ^*]

we havn't learnt this yet...but i want a get a head start (since i'm not as gifted as other people in the class )

could someone plz show me the steps involved in answering this question...

 

[:: ck ::]

try t formuula

 

[ToO LaZy ^*]

is the answer..

blehh...i think its wrong

 

[kimmeh]

Using ryan's suggestion:

umm i scanned it funny.. so from right to left

hopefully its not wrong

 

[ToO LaZy ^*]

cheeers!!
thx kimmeh

 

[kimmeh]

whered you get ur avatar from? its cute . cro_angel had one that was similar

 

[ToO LaZy ^*]

err..i can't remember precisely, i googled it under 'funny avatars' and i found it somewhere on the first page

 

[meyero]

Simple

from the standard integral sheet it looks like Arctan(e^-x) to me.