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How do you find the antiderivative of #(e^(2x))/(1+e^(2x))#?

Answer 1

Ezra Adams

# int \ e^(2x)/(1+e^(2x)) \ dx = 1/2ln|1+e^(2x)| + c#

We want to find # int \ e^(2x)/(1+e^(2x)) \ dx #

A trivial substitution can be used to simplify the denominator; Let:

# u = 1+e^(2x) #

Then # (du)/dx = 2e^(2x) => 1/2 (du)/dx = e^(2x)#

If we substitute this into the integral we get;

# int \ e^(2x)/(1+e^(2x)) \ dx = int \ (1/2)/(u) \ du# # " " = 1/2ln|u| + c#

And if we undo the substitution we get:

# int \ e^(2x)/(1+e^(2x)) \ dx = 1/2ln|1+e^(2x)| + c#

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Answer 2

Axel Alvarez

To find the antiderivative of (\frac{e^{2x}}{1+e^{2x}}), you can use the substitution method. Let (u = 1 + e^{2x}), then (du = 2e^{2x}dx). Rearrange to solve for (dx), getting (dx = \frac{du}{2e^{2x}}). Now substitute into the integral. This gives:

[ \int \frac{e^{2x}}{1+e^{2x}} , dx = \int \frac{1}{u} \cdot \frac{du}{2e^{2x}} = \frac{1}{2} \int \frac{1}{u} , du ]

This simplifies to:

[ \frac{1}{2} \ln|u| + C = \frac{1}{2} \ln|1 + e^{2x}| + C ]

where (C) is the constant of integration.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.