# How do you integrate int ln(3x) by parts?

Mar 12, 2017

$\int \ln \left(3 x\right) \mathrm{dx} = x \left(\ln \left(3 x\right) - 1\right) + C$

#### Explanation:

Integration by parts tells us that:

$\int u \left(x\right) v ' \left(x\right) \mathrm{dx} = u \left(x\right) v \left(x\right) - \int v \left(x\right) u ' \left(x\right) \mathrm{dx}$

In our example, put:

$\left\{\begin{matrix}u \left(x\right) = \ln \left(3 x\right) \\ v \left(x\right) = x\end{matrix}\right.$

Then:

$\left\{\begin{matrix}u ' \left(x\right) = 3 \cdot \frac{1}{3 x} = \frac{1}{x} \\ v ' \left(x\right) = 1\end{matrix}\right.$

So we find:

$\int \ln \left(3 x\right) \mathrm{dx} = \int u \left(x\right) v ' \left(x\right) \mathrm{dx}$

$\textcolor{w h i t e}{\int \ln \left(3 x\right) \mathrm{dx}} = u \left(x\right) v \left(x\right) - \int v \left(x\right) u ' \left(x\right) \mathrm{dx}$

$\textcolor{w h i t e}{\int \ln \left(3 x\right) \mathrm{dx}} = x \ln \left(3 x\right) - \int x \cdot \frac{1}{x} \mathrm{dx}$

$\textcolor{w h i t e}{\int \ln \left(3 x\right) \mathrm{dx}} = x \ln \left(3 x\right) - \int 1 \mathrm{dx}$

$\textcolor{w h i t e}{\int \ln \left(3 x\right) \mathrm{dx}} = x \ln \left(3 x\right) - x + C$

$\textcolor{w h i t e}{\int \ln \left(3 x\right) \mathrm{dx}} = x \left(\ln \left(3 x\right) - 1\right) + C$

Mar 12, 2017

The answer is $= x \left(\ln \left(| 3 x |\right) - 1\right) + C$

#### Explanation:

$\int u ' v \mathrm{dx} = u v - \int u v ' \mathrm{dx}$

Let $v = \ln \left(3 x\right)$, $\implies$, $v ' = \frac{1}{3 x} \cdot 3 = \frac{1}{x}$

$u ' = 1$, $\implies$, $u = x$

Therefore,

$\int \ln \left(3 x\right) \mathrm{dx} = x \ln \left(3 x\right) - \int \frac{1}{x} \cdot x \mathrm{dx}$

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

$= x \left(\ln \left(3 x\right) - 1\right) + C$