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

Mar 7, 2018

$x \ln \left(3 x\right) - x + C$

#### Explanation:

We can do this in a number of ways.

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

For $\ln \left(x\right)$, we can write the following:
$\frac{\mathrm{dy}}{\mathrm{dx}} = \ln \left(x\right) \rightarrow \mathrm{dy} = \ln \left(x\right) \mathrm{dx}$
Let's use u substitution with $u = \ln \left(x\right)$, i.e. $\mathrm{du} = \frac{1}{x} \mathrm{dx} = {e}^{- u} \mathrm{dx}$. This yields

$\mathrm{dy} = u {e}^{u} \mathrm{du} \implies y = \int u {e}^{u} \mathrm{du}$
By integration by parts,
$y = u {e}^{u} - \int {e}^{u} \mathrm{du} = u {e}^{u} - {e}^{u} + C$
Putting back $x$,

$y = x \ln x - x + C$

This yields the final antiderivative as
$\int \setminus \ln \left(3 x\right) \mathrm{dx} = x \ln \left(3\right) + x \ln \left(x\right) - x + C = x \ln \left(3 x\right) - x + C$