# Evaluating integrals on exponential functions with natural logs.

• Mar 12th 2013, 08:21 PM
Chaim
Evaluating integrals on exponential functions with natural logs.
∫(e-x/(1+e-x))
So I used substituion, u = 1 + e-x, and -du = e-x dx
There it turns it into -∫(du/u), which equals -ln(1+e-x) + C
Then my partner told me the next step was ln(ex/(ex+1))
After that, he got it into x - ln (ex + 1) + C, which is the correct answer.
And I was confused on that part since it turned into something like a fraction. The part where it goes e x/ex+1 Can someone explain?

Another is ∫ex√(1-ex)dx
I did u = 1-ex, -du = ex dx
I plugged it in...
∫-du√(u) = -∫du(u1/2)
Not really sure if I'm doing this next step correctly... but I did that it = u1/2
Plugged in u: (1-ex)1/2

But the answer was: -2/3(1-ex)3/2 + C

I'm a bit confused now... since I'm integrating the 'e' along...

Thanks :)
• Mar 12th 2013, 10:47 PM
princeps
Re: Evaluating integrals on exponential functions with natural logs.
1. $\displaystyle -\ln\left(1+\frac{1}{e^x}\right)=-\ln\left(\frac{e^x+1}{e^x}\right)=-\ln\left(\frac{e^x}{e^x+1}\right)^{-1}=\ln \frac{e^x}{e^x+1}$

2. $\displaystyle \int \sqrt{u}\, du =\frac{u^{3/2}}{3/2}+C$
• Mar 13th 2013, 07:18 AM
Chaim
Re: Evaluating integrals on exponential functions with natural logs.
Quote:

Originally Posted by princeps
1. $\displaystyle -\ln\left(1+\frac{1}{e^x}\right)=-\ln\left(\frac{e^x+1}{e^x}\right)=-\ln\left(\frac{e^x}{e^x+1}\right)^{-1}=\ln \frac{e^x}{e^x+1}$

2. $\displaystyle \int \sqrt{u}\, du =\frac{u^{3/2}}{3/2}+C$

Oh wow.
Now that makes more sense!

I forgot to see that it had a negative exponent, which makes it goes down into the denominator, so the only way to convert the 1 to have them both at the same denominator is by having it ex as well!

While for the second one. I think I remember now! So basically after you integrate it, the exponent goes up by 1 and you would divide it by the exponent.

Thanks for helping me and reminding me of this stuff! :)