Integration question?

Question

How do I integrate:

1/((e^-y)+1) dy

Answer 1

Integral ( 1 / [e^(-y) + 1] ) dy

The first thing you need to do is express e^(-y) as 1/(e^y)

Integral (1 / [1/(e^y) + 1)]) dy

Notice we now have a complex fraction. We have to multiply top and bottom by e^y, giving us

Integral (e^y / (1 + e^y)) dy

Let me change this around a bit to make something obvious.

Integral ( [1/(1 + e^y)](e^y)dy )

Now, we use substitution.
Let u = 1 + e^y.
du = (e^y)dy

Now, we can replace the tail end of that integral with du, since "e^y dy" is there.

Integral (1/u)du

This is now easy to solve

ln|u| + C

And replacing u = 1 + e^x back, we have

ln |1 + e^x| + C

Answer 2

Make the change e^-y = z ==> -e^-y dy = dz and dy = -dz/z
So it comes to integrate -dz/(z+1)*(z) .You should use partial fraction (a/z +b(z+1))dz

Answer 3

Do it by partial fractions.

Answer 4

=[ln( e^-y +1)] e^-y