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High School
Calculus and Analysis
Calculus 2
Applications of integrals

hunterofdeath63

Answered question

2021-12-10

In the given equation as follows , use a table of integrals with forms involving the trigonometric functions to find the indefinite integral:

\int \frac{1}{1 + e^{2 x}} d x

Answer & Explanation

Bertha Jordan

Beginner2021-12-11Added 37 answers

Step 1
Let $1 + e^{2 x} = t$ . Differentiate both sides.
$e^{2 x} \times 2 d x = d t$
$d x = \frac{d t}{2 e^{2 x}}$
$= \frac{d t}{2 (t - 1)}$
Substitute the values into the integral. Also since the integral is indefinite; a constant of integration is to be added.
Step 2
Perform the integration.
$\int \frac{1}{1 + e^{2 x}} d x = \int \frac{1}{t} \times \frac{d t}{2 (t - 1)}$
$= \frac{1}{2} \int \frac{d t}{t (t - 1)}$
$= \frac{1}{2} \int [\frac{1}{t - 1} - \frac{1}{t}] d t$
$= \frac{1}{2} \times [\ln | t - 1 | - \ln | t |] + c$
$= \frac{1}{2} \times [\ln | 1 + e^{2 x} - 1 | - \ln | 1 + e^{2 x} |] + c$
$= \frac{1}{2} \times [\ln | e^{2 x} | - \ln | 1 + e^{2 x} |] + c$
Hence the solution is obtained.

SlabydouluS62

Skilled2021-12-12Added 52 answers

\int \frac{1}{e^{2 x} + 1} d x

= - \frac{1}{2} \int \frac{1}{e^{- u} + 1} d u

= \int \frac{e^{u}}{e^{u} + 1} d u

= \int \frac{1}{v} d v

= \ln (v)

= \ln (e^{u} + 1)

- \frac{1}{2} \int \frac{1}{e^{- u} + 1} d u

= - \frac{\ln (e^{u} + 1)}{2}

Answer:

= - \frac{\ln (e^{u} + 1)}{2} + C

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