Anne Wacker

2021-12-28

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

### Answer & Explanation

Paineow

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 $\int \frac{{e}^{x}-{e}^{-x}}{{e}^{x}+{e}^{-x}}dx$. Consider denominator of integrand ${e}^{x}+{e}^{-x}$ as u. Differentiate the denominator with respect to x
${e}^{x}+{e}^{-x}=u$
$\frac{d\left({e}^{x}+{e}^{-x}\right)}{dx}=\frac{du}{dx}$
${e}^{x}-{e}^{-x}=\frac{du}{dx}$
$du=\left({e}^{x}-{e}^{-x}\right)dx$...(1)
Step 3
Substitute the value of $\left({e}^{x}-{e}^{-x}\right)dx=du$ and ${e}^{x}+{e}^{-x}=u$ in the integrand $\int \frac{{e}^{x}-{e}^{-x}}{{e}^{x}+{e}^{-x}}dx$
$\int \frac{{e}^{x}-{e}^{-x}}{{e}^{x}+{e}^{-x}}=\int \frac{1}{u}du$
$=\mathrm{ln}|u|+C$
$=\mathrm{ln}|{e}^{x}+{e}^{-x}|+C$ (As $u={e}^{x}+{e}^{-x}$)
Therefore, integration of $\int \frac{{e}^{x}-{e}^{-x}}{{e}^{x}+{e}^{-x}}dx$ is $\mathrm{ln}|{e}^{x}+{e}^{-x}|+C$

puhnut1m

It is required to calculate:
$\int \frac{{e}^{x}-{e}^{-x}}{{e}^{x}+{e}^{-x}}dx$
Substitution $u={e}^{x}+{e}^{-x}⇒\frac{du}{dx}={e}^{x}-{e}^{-x}⇒dx=\frac{1}{{e}^{x}-{e}^{-x}}du$
$=\int \frac{1}{u}du$
This is the well-known tabular integral:
$=\mathrm{ln}\left(u\right)$
Reverse replacement $u={e}^{x}+{e}^{-x}:$
$=\mathrm{ln}\left({e}^{x}+{e}^{-x}\right)$
$=\mathrm{ln}\left({e}^{x}+{e}^{-x}\right)+C$

Vasquez

$\begin{array}{}\int \frac{{e}^{x}-{e}^{-x}}{{e}^{x}+{e}^{-x}}\\ \int \frac{1}{t}dt\\ \mathrm{ln}\left(|t|\right)\\ \mathrm{ln}\left(|{e}^{x}+{e}^{-x}|\right)\\ \mathrm{ln}\left({e}^{2x}+1\right)-x\\ \text{Solution:}\\ \mathrm{ln}\left({e}^{2x}+1\right)-x+C\end{array}$