How do you evaluate the integral #inte^(4x) dx#?

1 Answer
Aug 13, 2014

We will use #u#-substitution, letting #u = 4x#.

Thus, #du = 4dx#.

Also, we will use the constant law of integration, namely #int C*f(x)dx = C*int f(x) dx# to rewrite the integral so that it contains #du#:

#int e^(4x)dx = 1/4 int 4*e^(4x)dx#

Now, we will rewrite in terms of #u#:

#int e^(4x)dx = 1/4 int e^(u)du#

We know that the integral of #e^u du# will simply be #e^u#. Remember the constant of integration:

#int e^(4x)dx = 1/4 e^(u) + C#

Substituting back #u# gives:

#int e^(4x)dx = 1/4 e^(4x) + C#