Question

Help please;

Integrate following :
1/(e^x + 1)

Thanks in advance.

Answer 1

Rewrite it as:
\[\int\limits \frac{1+e^x-e^x}{e^x+1} dx=\int\limits \frac{e^x+1}{e^x+1}dx-\int\limits \frac{e^x}{e^x+1} dx\].
The first integral reduces to the integral of 1. The second is a u-sub.

Answer 2

ln(e^x(e^x+1))+c

Answer 3

Rewrite it as:
\[\int\limits \frac{1+e^x-e^x}{e^x+1} dx=\int\limits \frac{e^x+1}{e^x+1}dx-\int\limits \frac{e^x}{e^x+1} dx\].
The first integral reduces to the integral of 1. The second is a u-sub.

Answer 4

let p = e^x+1.
So integrand becomes 1/p.
Integral is ln(p) + c.
is this right?

Answer 5

no. it seems wrong.

Answer 6

Let e^x=t
e^x dx = dt
dx=dt/e^x=dt/t
Hence the integral is dt/(t(t+1))
=dt/(t^2+t)=dt/((t+1/2)^2-1/4)
let t+1/2=k
the integral now is dk/(k^2-1/4)
=ln(k-1/2)-ln(k+1/2)=ln(t)-ln(t+1)=ln(e^x)-ln(e^x+1)= x-ln(e^x+1)

Answer 7

Damn shankvee.

Answer 8

pls wait. let me go thru it.

Answer 9

kiranbandal whenever you substitute you have to care care of dx. in your case dp=e^x dx

Answer 10

malevolence what did i do? I just gave a method to integrate the function which is a pretty simple one.

Answer 11

I'm just saying :P double substitution? haha.

Answer 12

Shankvee you changed the variables and did more than one substitution ... That was way more complicated than it needed to be

Answer 13

and Kiranbandal, you can also multiply top and bottom by e^(-x) it will also simplify it into less work than my previous post.

Answer 14

You still get the answer dont you? There is not much complixation You should get used to it if you practice.

Answer 15

ok, let me try "multiply top and bottom by e^(-x)"

Answer 16

I mean, I'm not knocking your approach. Just seems like a lot work xP

Answer 17

malevolence what is your problem with double substitutions? :D

Answer 18

(e^x+1) e^-x ... what would it be?

Answer 19

I suggest complicating it more by integrating 1/(t(t+1)) by parital fractions :P

Answer 20

ok. i got it. thanks shankvee and malevolence19.

Answer 21

my answer is: - ln | e^-x + 1 | + C.
I divided top and bottom by e^x. So, i got f'(x) / f(x).

Answer 22

actually when you nultiply and divide e^x you will have e^x in the denominator also so differential of e^x(1+e^x) is not equal to e^x

Answer 23

so, is my answer wrong?

Answer 24

unfortunately yes. :D

Answer 25

see, d/dx of (e^-x + 1) is -e^x. so, i introduced -1 inside and outside of integrand.

Answer 26

Shank-I don't have a problem with it :P Just not good for rookies xDDD