Anne Wacker

2021-12-28

Find the indefinite integral.

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

Paineow

Beginner2021-12-29Added 30 answers

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

${e}^{x}+{e}^{-x}=u$

$\frac{d({e}^{x}+{e}^{-x})}{dx}=\frac{du}{dx}$

$e}^{x}-{e}^{-x}=\frac{du}{dx$

$du=({e}^{x}-{e}^{-x})dx$ ...(1)

Step 3

Substitute the value of$({e}^{x}-{e}^{-x})dx=du$ and ${e}^{x}+{e}^{-x}=u$ in the integrand $\int \frac{{e}^{x}-{e}^{-x}}{{e}^{x}+{e}^{-x}}dx$

$\int \frac{{e}^{x}-{e}^{-x}}{{e}^{x}+{e}^{-x}}=\int \frac{1}{u}du$

$=\mathrm{ln}\left|u\right|+C$

$=\mathrm{ln}|{e}^{x}+{e}^{-x}|+C$ (As $u={e}^{x}+{e}^{-x}$ )

Therefore, integration of$\int \frac{{e}^{x}-{e}^{-x}}{{e}^{x}+{e}^{-x}}dx$ is $\mathrm{ln}|{e}^{x}+{e}^{-x}|+C$

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

Step 3

Substitute the value of

Therefore, integration of

puhnut1m

Beginner2021-12-30Added 33 answers

It is required to calculate:

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

Substitution$u={e}^{x}+{e}^{-x}\Rightarrow \frac{du}{dx}={e}^{x}-{e}^{-x}\Rightarrow dx=\frac{1}{{e}^{x}-{e}^{-x}}du$

$=\int \frac{1}{u}du$

This is the well-known tabular integral:

$=\mathrm{ln}\left(u\right)$

Reverse replacement$u={e}^{x}+{e}^{-x}:$

$=\mathrm{ln}({e}^{x}+{e}^{-x})$

Answer:

$=\mathrm{ln}({e}^{x}+{e}^{-x})+C$

Substitution

This is the well-known tabular integral:

Reverse replacement

Answer:

Vasquez

Expert2022-01-07Added 669 answers

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