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 l__n(e__^{x}/(e^{x+1)})

After that, he got it into __x - ln (e__^{x} + 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}/e^{x+1} Can someone explain?

Another is ∫e^{x}√(1-e^{x})dx

I did u = 1-e^{x}, -du = e^{x} dx

I plugged it in...

∫-du√(u) = -∫du(u^{1/2})

Not really sure if I'm doing this next step correctly... but I did that it = u^{1/2}

Plugged in u: (1-e^{x})^{1/2}

But the answer was: -2/3(1-e^{x})^{3/2} + C

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

Can someone give a piece of advice or explain please?

Thanks :)

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

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 e^{x} 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! :)