Expert answers to "Find the indefinite integral. ∫e2x1+e2xdx"

rheisf

Answered question

2021-12-13

Find the indefinite integral.

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

Jim Hunt

Beginner2021-12-14Added 45 answers

Step 1 : Analysis
Given :

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

Step 2 : Simplification
Let the integral

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

Substituting

u = 1 + e^{2 x}

\frac{d u}{d x} = \frac{d}{d x} (1 + e^{2 x})

\frac{d u}{d x} = \frac{d}{d x} (1) + \frac{d}{d x} (e^{2 x})

\frac{d u}{d x} = 0 + e^{2 x} \frac{d}{d x} (2 x)

\frac{d u}{d x} = 2 e^{2 x}

\frac{d u}{2 e^{2 x}} = d x

Step 3 : Solution
We have,

I = \int \frac{e^{2 x}}{u} \times \frac{d u}{2 e^{2 x}}

I = \frac{1}{2} \int \frac{d u}{u}

I = \frac{1}{2} \ln | u | + C

Therefore the integral is

\int \frac{e^{2 x}}{1 + e^{2 x}} d x = \frac{1}{2} \ln | 1 + e^{2 x} | + C

where C is the integral constant.

Rita Miller

Beginner2021-12-15Added 28 answers

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

Let

u = 1 + e^{2 x} \Rightarrow d u = 2 e^{2 x} d x

\int \frac{e^{2 x}}{1 + e^{2 x}} d x = \int \frac{\frac{1}{2} d u}{u} = \frac{1}{2} \ln (u) + c = \frac{1}{2} \ln (1 + e^{2 x}) + C

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