Expert: Paul Klarreich - 1/27/2010

Question
1. Find the following integral. Note that you can check your answer by differentiation.

{
| (e^(4x))/(2+e^(4x))
}

For this I've tried a few different ways and seem to not get the answer. This is what I was trying.

let u=4x <-> x= u/4
           (1/4)du=dx
    {
(1/4)| (e^(4x))/(2+e^(4x)) dx
    }

    {
(1/4)| (e^u)/(2+e^(u)) du
    }

then i get lost here.. i don't even know if i'm doing it right up to what i did get.how do I do this using substitution?
Mike


Also, there are some problems I am running into when doing substitution for instance when you have something like this,
let u=2x
  du=2??
how do you go about that? is there another way? like for instance it is an indefinite integral where the limits are not given where if it was given you would plug in the limits into u=2x and get the newer limits (if I am even correct) but i keep running into problems where du= is left without variables. just wondering.thanks.  

Answer
Questioner: Mike
Country: United States
Category: Calculus
Private: Yes
Subject: Indefinite Integrals using Substitution
Question: 1. Find the following integral. Note that you can check your answer by differentiation.

{
| (e^(4x))/(2+e^(4x))  TIMES dx, please
}

For this I've tried a few different ways and seem to not get the answer. This is what I was trying.

let u=4x <-> x= u/4
          (1/4)du=dx
   {
(1/4)| (e^(4x))/(2+e^(4x)) dx
   }

   {
(1/4)| (e^u)/(2+e^(u)) du
   }

then i get lost here.. i don't even know if i'm doing it right up to what i did get.how do I do this using substitution?

......................................
Two things:

1. If your u-subst is not working out, go back and try something different.

2. If your u-subst is not working out, maybe you have more work to do.

In this case, you have:

{      e^u du
| --------------
} 4 (2 + e^u)

Now do more:  let  v = e^u,  dv = e^u du, and you have:

{      dv
| ----------
} 4 (2 + v)

which you can handle as a ln.

You could even do it as:

let  w = 2 + e^u,  dw = e^u du, and you have:

{  dw
| -----
} 4 w

even easier.

OR YOU COULD HAVE DONE IT ALL IN ONE SHOT FROM HERE:

{   e^(4x)dx
|   ------------
}   2+e^(4x)

let  u = 2 + e^4x,  du = e^4x dx

{   du
| -----
}    u

............................
Mike

Also, there are some problems I am running into when doing substitution for instance when you have something like this,
let u=2x
 du=2??
how do you go about that? is there another way? like for instance it is an indefinite integral where the limits are not given where if it was given you would plug in the limits into u=2x and get the newer limits (if I am even correct) but i keep running into problems where du= is left without variables. just wondering.thanks.

If you have a definite integral, you are 'supposed to' change the boundaries when you change the variable, but unless there is a good reason to, I don't know anyone who does it.  Everyone just re-substitutes at the end and uses the original values.