# How do you integrate int ln 4x dx  using integration by parts?

Apr 6, 2018

$= x \ln 4 x - x + C$

#### Explanation:

$\int \ln 4 x \mathrm{dx}$

You can't integrate $\ln 4 x$ so you pick

$u = \ln 4 x$ $\rightarrow$ $\mathrm{du} = \frac{1}{x} \mathrm{dx}$
$\mathrm{dv} = \mathrm{dx}$$\rightarrow$$v = x$

And by using the formula:
$\int u \mathrm{dv} = u v - \int v \mathrm{du}$

$\int \ln 4 x \mathrm{dx} = x \ln 4 x - \int \frac{x}{x} \mathrm{dx}$

$= x \ln 4 x - \int 1 \mathrm{dx}$

$= x \ln 4 x - x + C$

Apr 6, 2018

The answer is $= x \left(\ln \left(| 4 x |\right) - 1\right) + C$

#### Explanation:

Perform the integration by parts

$\int u v ' = u v - \int u ' v$

Here,

$u = \ln 4 x$, $\implies$, $u ' = \frac{1}{4 x} \cdot 4 = \frac{1}{x}$

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

Therefore,

$\int \ln 4 x = x \ln 4 x - \int \frac{1}{x} \cdot x \mathrm{dx}$

$= x \ln 4 x - x + C$

$= x \left(\ln \left(| 4 x |\right) - 1\right) + C$