Do you have a diferent method to solve this Integral?

[Bryan1]

Integral[(dx/(e^x+1)]
Here’s my method:
If
x = -Ln(u)
Then
dx = -du/u;
e^x = 1/u;
u = e^-x;
Substituting in the integral we get:
Integral[(-du/u)/((1/u)+1)] = -Integral[du/(1+u)]
Solving the integer in terms of u:
-Ln|1+u|+C
Substituting x in the result:
-Ln|1+e^-x|+C
Sorry that i didn't use the appropriate math symbols, I don't unsderstand the language used to do it. And also sorry if there's english mistakes.
Tank you for reading up to this point.

 

[MarkFL]

I would write:

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

 

[Prove It]

Integral[(dx/(e^x+1)]
Here’s my method:
If
x = -Ln(u)
Then
dx = -du/u;
e^x = 1/u;
u = e^-x;
Substituting in the integral we get:
Integral[(-du/u)/((1/u)+1)] = -Integral[du/(1+u)]
Solving the integer in terms of u:
-Ln|1+u|+C
Substituting x in the result:
-Ln|1+e^-x|+C
Sorry that i didn't use the appropriate math symbols, I don't unsderstand the language used to do it. And also sorry if there's english mistakes.
Tank you for reading up to this point.

$\displaystyle \begin{align*} \int{\frac{dx}{e^x + 1}} &= \int{\frac{e^x\,dx}{e^x \left( e^x + 1 \right)} } \\ &= \int{ \frac{du}{u \left( u +1 \right) } } \textrm{ if we let } u = e^x \implies du = e^x\,dx \end{align*}$

So now applying partial fractions:

$\displaystyle \begin{align*} \frac{A}{u} + \frac{B}{u + 1} &\equiv \frac{1}{u \left( u + 1 \right) } \\ A \left( u + 1 \right) + B \, u &\equiv 1 \end{align*}$

Let $\displaystyle \begin{align*} u = 0 \end{align*}$ and we find $\displaystyle \begin{align*} A = 1 \end{align*}$ and let $\displaystyle \begin{align*} u = -1 \end{align*}$ to find $\displaystyle \begin{align*} B = -1 \end{align*}$ and we have

$\displaystyle \begin{align*} \int{ \frac{du}{u \left( u + 1 \right) } } &= \int{ \frac{1}{u} - \frac{1}{u + 1} \,du } \\ &= \ln{ \left| u \right| } - \ln{ \left| u + 1 \right| } + C \\ &= \ln{ \left| e^x \right| } - \ln{ \left| e^x + 1 \right| } + C \\ &= \ln{ \left( e^x \right) } - \ln{ \left( e^x + 1 \right) } + C \textrm{ since } e^x > 0 \textrm{ for all } x \\ &= x - \ln{ \left( e^x + 1 \right) }+ C \end{align*}$