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

Answer 1

#xln(3x) - x + C#

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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Answer 2

Ezra Adams

The antiderivative of ln(3x) is (1/3) * (3x) * ln(3x) - x + 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.