What is the antiderivative of # ln(3x)#?

1 Answer
Jake M.
Mar 7, 2018

#xln(3x) - x + C#

Explanation:

We can do this in a number of ways. 

First, let's get rid of that three using a log property:
#ln(3x) = ln(3) + ln(x) #
ln(3) is a constant, so its antiderivative will just be #xln(3)# which is fine to deal with later.

For #ln(x)#, we can write the following:
#dy/dx = ln(x) rightarrow dy = ln(x) dx#
Let's use u substitution with #u=ln(x)#, i.e. #du = 1/x dx = e^(-u) dx#. This yields

#dy = u e^u du implies y = int u e^u du #
By integration by parts,
#y = ue^u - int e^u du = ue^u - e^u + C #
Putting back #x#,

#y = xlnx - x + C #

This yields the final antiderivative as
#int\ ln(3x)dx = xln(3) + xln(x) - x + C = xln(3x) - x + C #

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