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How do you integrate #int e^(2x)/(1+e^(2x))dx#?

Answer 1

Elijah Abram

The answer is #=1/2ln(1+e^(2x))+C#

Let's do it by substitution

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

then, #du=2e^(2x)dx#

#int(e^(2x)dx)/(1+e^(2x))#

#=1/2int(du)/u=1/2lnabsu#

#=1/2ln(1+e^(2x))+C#

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Answer 2

Charles Anctil

To integrate ( \int \frac{e^{2x}}{1+e^{2x}} , dx ), we can use a substitution method. Let ( u = 1 + e^{2x} ). Then, ( du = 2e^{2x} , dx ). Solving for ( dx ), we get ( dx = \frac{du}{2e^{2x}} ). Substituting these into the integral 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 integral is straightforward to solve. After integrating, we resubstitute ( u ) back in terms of ( x ) to get the final result.

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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.