1.
2.
∫(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 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
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!