# Thread: Help with Natural exponential integrals

1. ## Help with Natural exponential integrals

Hi all!
im stuck again with this difficult integral

$\int \frac{dx}{1+e^x}$

can u help with what to substitute
??

and we're not yet in integration by parts

2. Originally Posted by ^_^Engineer_Adam^_^
Hi all!
im stuck again with this difficult integral

$\int \frac{dx}{1+e^x}$

can u help with what to substitute
??

and we're not yet in integration by parts
Try the substitution $u=e^x$, then the integral becomes:

$
\int \frac{1}{u(1+u)} du = \int \frac{1}{u}-\frac{1}{1+u}du
$

RonL

3. Here is a second way:
$\int {\frac{{dx}}{{1 + e^x }}} = \int {\frac{{e^{ - x} dx}}{{e^{ - x} + 1}} = - \ln \left( {e^{ - x} + 1} \right)}.$

4. Originally Posted by Plato
Here is a second way:
$\int {\frac{{dx}}{{1 + e^x }}} = \int {\frac{{e^{ - x} dx}}{{e^{ - x} + 1}} = - \ln \left( {e^{ - x} + 1} \right)}.$
I think you forgot something

5. Originally Posted by ^_^Engineer_Adam^_^
can u help with what to substitute
??

and we're not yet in integration by parts
I suggest when you post a problem, expose your solution or part of it to see the progress and possibles mistakes that you could have.

Regards

6. Originally Posted by ^_^Engineer_Adam^_^
Hi all!
im stuck again with this difficult integral

$\int \frac{dx}{1+e^x}$

can u help with what to substitute
??

and we're not yet in integration by parts
Yet another possibility:

$\int \frac {1}{1 + e^x}~dx = \int \frac {1 + e^x - e^x}{1 + e^x}$

$= \int \left( 1 - \frac {e^x }{1 + e^x} \right)~dx$

$= x - \ln |1 + e^x| + C$

The possibilities are endless

7. Originally Posted by Jhevon
Yet another possibility:

$\int \frac {1}{1 + e^x}~dx = \int \frac {1 + e^x - e^x}{1 + e^x}$

$= \int \left( 1 - \frac {e^x }{1 + e^x} \right)~dx$

$= x - \ln |1 + e^x| + C$

The possibilities are endless
I like this solution.

This is one of my favorite integrals.
I posted a long discourse on it on SOSMath in 2003.

$\int \frac{dx}{1+e^x}$
(1) Plato had the best method: .multiply top and bottom by $e^{-x}$
His answer: . $-\ln(e^{-x} +1) + C$

(2) CaptainBlack also had a good approach: Partial Fractions.

He had: . $\int\left(\frac{1}{u} - \frac{1}{u+1}\right)\,du \;=\;\ln(u) - \ln(u + 1) + C$

Back-substitute: . $\ln(e^x) - \ln(e^x + 1) + C \;=\;x - \ln(e^x + 1) + C$

(3) We can use Long Division:. . $\int\frac{1}{1+e^x}\,dx \;=\;\int\frac{1 + e^x - e^x}{1 + e^x}\,dx \;=\;\int\left(\frac{1+e^x}{1+e^x} - \frac{e^x}{1+e^x}\right)\,dx$
. . $=\;\int\left(1 - \frac{e^x}{1 + e^x}\right)\,dx \;= \;x - \ln(1 + e^x) + C$

(4) One day, while totally bored, I came up with this substitution . . .

Let $e^x = \tan^2\!\theta \quad\Rightarrow\quad e^{\frac{x}{2}} = \tan\theta \quad\Rightarrow\quad \frac{x}{2} = \ln(\tan\theta) \quad\Rightarrow\quad dx = 2\frac{\sec^2\theta}{\tan\theta}\,d\theta$
Substitute: . $\int\frac{2\frac{\sec^2\theta}{\tan\theta}\,d\thet a}{\sec^2\theta} \;=\;2\int\cot\theta\,d\theta \;=\;2\ln(\sin\theta) + C$

Since $\tan\theta = e^{\frac{x}{2}}$, then $\theta$ is in a right triangle
. . with: . $opp = e^{\frac{x}{2}}$ and $adj = 1$
Hence: . $hyp = \sqrt{1 + e^x}$ . . . and: . $\sin\theta = \frac{e^{\frac{x}{2}}}{\sqrt{1 + e^x}}$
And we have: . $2\ln\left(\frac{e^{\frac{x}{2}}}{\sqrt{1+e^x}}\rig ht) + C$

And, of course, you are expected to show that these answers are equivalent.

9. Well if you really want to be ugly I have another way of doing this.

$\frac{1}{1+e^x} = \frac{1}{2+x+\frac{x^2}{2!}+\frac{x^3}{3!}+....}$

Long division.

Integrate term-by-term

And bring to standard form of ones of the know functions.

I leave the details to you.

10. Originally Posted by ThePerfectHacker
$\frac{1}{1+e^x} = \frac{1}{3+x+\frac{x^2}{2!}+\frac{x^3}{3!}+....}$
Where does the "3" come in the denominator?
It should be "2" I suppose.

11. Originally Posted by curvature
Where does the "3" come in the denominator?
It should be "2" I suppose.
Yes I fixed it.