jamessinatraaa

2021-12-27

Find the indefinite integral.
$\int \frac{{e}^{-x}}{1+{e}^{-x}}dx$

lalilulelo2k3eq

Step 1
Integration is summation of discrete data. The integral is calculated for the functions to find their area, displacement, volume, that occurs due to combination of small data.
Integration is of two types definite integral and indefinite integral. Indefinite integral are defined where upper and lower limits are not given, whereas in definite integral both upper and lower limit are there.
Step 2
The given integrand is $\int \frac{{e}^{-x}}{1+{e}^{-x}}dx$. Consider the denominator of the integrand $1+{e}^{-x}$, let it be equal to u. Differentiate the denominator of integrand with respect to x .
$u=1+{e}^{-x}$
$\frac{du}{dx}=\frac{d\left(1={e}^{-x}\right)}{dx}$
$=\frac{d1}{dx}+\frac{d{e}^{-x}}{dx}$
$=0+{e}^{-x}\left(-1\right)$
$du=-1\left({e}^{-x}\right)dx$
$\left(-1\right)du={e}^{-x}dx$...(1)
Step 3
Substitute value of $u=1+{e}^{-x}$ and ${e}^{-x}dx=\left(-1\right)du$ from equation (1) in $\int \frac{{e}^{-x}}{1+{e}^{-x}}dx$
$\int \frac{{e}^{-x}}{1+{e}^{-x}}dx=\int \frac{\left(-1\right)du}{u}$
$-1\mathrm{ln}|u|+C$
$=-1\mathrm{ln}|1+{e}^{-x}|+C$ (As $1+{e}^{-x}=u$)
Therefore, integration of $\int \frac{{e}^{-x}}{1+{e}^{-x}}dx$ is $-1\mathrm{ln}|1={e}^{-x}|+C$

Mary Nicholson

$\int \frac{{e}^{-x}}{1+{e}^{-x}}dx$
We put the expression exp (x) under the sign of the differential, i.e.:
${e}^{x}dx=d\left({e}^{x}\right),t={e}^{x}$
Then the original integral can be written as follows: We put the
$\int \frac{1}{{t}^{2}\cdot \left(1+\frac{1}{t}\right)}dt$
Expression $-\frac{1}{{x}^{2}}$ under the sign of the differential, i.e. ie.:
$\left(-\frac{1}{{x}^{2}}\right)dx=d\left(\frac{1}{x}\right),u=\frac{1}{x}$
Then the original integral can be written as follows:
$\int \left(-\frac{1}{u+1}\right)du$
$\int \frac{-1}{x+1}dx$
Calculate the tabular integral:
$-\int \frac{1}{x+1}dx=-\mathrm{ln}\left(x+1\right)$
$-\mathrm{ln}\left(x+1\right)+C$
Since previously we made a change of variables, then instead of u we substitute 1 / t.
$-\mathrm{ln}\left(1+\frac{1}{t}\right)+C$
To write down the final answer, it remains to substitute exp (x) instead of t.
$-\mathrm{ln}\left(1+{e}^{-x}\right)+C$

karton

$\begin{array}{}\int \frac{{e}^{-x}}{1+{e}^{-x}}dx\\ \int -\frac{1}{t}dt\\ -\int \frac{1}{t}dt\\ -\mathrm{ln}\left(|t|\right)\\ -\mathrm{ln}\left(|1+{e}^{-x}|\right)\\ -\mathrm{ln}\left(1+\frac{1}{{e}^{x}}\right)\end{array}$
$-\mathrm{ln}\left(1+\frac{1}{{e}^{x}}\right)+C$