How do you find the antiderivative of #e^(2x) / (3+e^(2x))#?

Question

Christopher Agresta · Accepted Answer

# int e^(2x) / (3+e^(2x)) \ dx = ln sqrt{ 3 + e^{2x)) + C#

easiest way always is to recognise the patterns

generalisation # d/(dx) ln (f(x)) = ( f'(x) ) / f(x) #

so if we consider # d/(dx) ln (3 + e^{2x}) = 1/(3 + e^{2x}) * 2 e^{2x}# then we're pretty much done

because # d/(dx) ln (3 + e^{2x}) = ( 2 e^{2x})/(3 + e^{2x}) # then we actually want # 1/2 * d/(dx) ln (3 + e^{2x}) = d/(dx) [ 1/2 * ln (3 + e^{2x})] # moving the constant inside the derivative

#= d/(dx) [ ln sqrt{ 3 + e^{2x)) \ ]#

thusly

# int e^(2x) / (3+e^(2x)) \ dx = ln sqrt{ 3 + e^{2x)) + C#

you can plough through a whole series of subs but seeing the pattern is a real life saver.

Charles Anctil · Answer

undefined To find the antiderivative of $ \frac{e^{2x}}{3 + e^{2x}} $, perform a substitution to simplify the expression.

Let $ u = e^{2x} $.

Then, $ du = 2e^{2x} dx $.

Now, rewrite the integral using the substitution:

$$ \int \frac{e^{2x}}{3 + e^{2x}} dx = \int \frac{1}{3 + u} \cdot \frac{du}{2} $$

This integral is now in a form that can be solved using a simple u-substitution:

$$ \int \frac{1}{3 + u} \cdot \frac{du}{2} = \frac{1}{2} \int \frac{du}{3 + u} $$

Integrate with respect to u:

$$ \frac{1}{2} \ln|3 + u| + C $$

Substitute back for u:

$$ \frac{1}{2} \ln|3 + e^{2x}| + C $$