∫(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(e^{x}/(e^{x+1)})

After that, he got it intox - 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