Logarithmic Integral Homework: Solving S [(e^x)(e^x)]/((2+(e^x))^2) dx

ppkjref
Messages
18
Reaction score
0

Homework Statement


S = integral symbol
S (e^(2x))/((2+(e^x))^2) dx


Homework Equations


u = 2+(e^x)
(e^x) = u-2
du = (e^x) dx


The Attempt at a Solution


S [(e^x)(e^x)]/((2+(e^x))^2) dx
S [du(u-2)]/(u^2)
For the first (e^x) dx I substituted du. And since there was only (e^x) left, I substituted in u-2.
Now what do I do?
I know how to do du/(u^2). but what about the (u-2)?
 
Physics news on Phys.org
So you need to find \int \frac{u-2}{u^2} du . Just split the numerator like this: \frac{a+b}{c} = \frac{a}{c} + \frac{b}{c}.
 
the a/c fraction is u/(u^2), equivalent to 1/u which when integrated is 0?
 
\int \frac{1}{u} du = \log_e u.

You should be familiar with that already.
 
ln(2+(e^x) + 2/(2+(e^x))
Just needed a break I guess. Wasn't thinking.
 
Last edited:
ppkjref said:
ln(2+(e^x)) + 2/(2+(e^x))
Just needed a break I guess. Wasn't thinking.

An unmatched left parentheses can haunt you all day.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top