# Integral of 1/2^x

1. Oct 25, 2015

1. The problem statement, all variables and given/known data
1 1/2x

2. Relevant equations

2-x = (eln 2)-x

Could possibly be relevant, I don't know.

3. The attempt at a solution

y = 2-x

ln y = -x ln 2

1/y dy/dx = -ln 2 -x/2

...

2. Oct 25, 2015

### HallsofIvy

Staff Emeritus
Better is $2^{-x}= e^{ln(2^{-x})}= e^{x ln(2)}$

3. Oct 25, 2015

### Samy_A

Quite relevant, try to build on that.

ETA: never mind, didn't see the previous post.

Last edited: Oct 25, 2015
4. Oct 25, 2015

∫(1/2x) dx = ∫(1/ex ln 2) dx

u = ln 2 x → du = ln 2 dx

1/ln 2 ∫ du/eu = ln eu / ln 2 = ln ex ln 2 / ln 2 = x ln 2/ln2 = x

Okay, what did I do wrong?

Edit: I was thinking of u'/ln u, my bad!

1/ln 2 ∫ e-u du = -e-u/ln 2 = -e-x ln 2/ln 2 = -2-x/ln2

Is this correct?

Last edited: Oct 25, 2015
5. Oct 25, 2015

Wouldn't that be e-x ln 2?

6. Oct 25, 2015

### Samy_A

You lost me in this last equality, I guess here something is wrong.

Also, don't forget to adapt the limits of your definite integral when you apply the substitution (not that I think that you really need that substitution).

7. Oct 25, 2015

### Samy_A

Yes, this is correct for the indefinite integral.

8. Oct 25, 2015

Yes I made an edit on my post, I was thinking of u'/ln u. I think I've got it now.

9. Oct 26, 2015

### HallsofIvy

Staff Emeritus
Yes. Thanks.