# What is int ln(2x)^2dx?

#### Answer:

$= 2 x \setminus \ln \left(2 x\right) - 2 x + C$

#### Explanation:

Given that

$\setminus \int \ln {\left(2 x\right)}^{2} \setminus \setminus d x$

$= \setminus \int 2 \ln \left(2 x\right) \setminus \setminus d x$

$= 2 \setminus \int \left(\ln x + \setminus \ln 2\right) \setminus \setminus d x$

$= 2 \setminus \int \ln x \setminus \mathrm{dx} + 2 \setminus \int \setminus \ln 2 \setminus d x$

$= 2 \left(\setminus \ln x \setminus \int 1 \setminus \mathrm{dx} - \setminus \int \left(\frac{1}{\mathrm{dx}} \left(\ln x\right) \setminus \cdot \setminus \int 1 \setminus \mathrm{dx}\right) \mathrm{dx}\right) + 2 \setminus \ln 2 \setminus \int 1 \setminus d x$

$= 2 \left(x \setminus \ln x - \setminus \int \left(\frac{1}{x} \setminus \cdot x\right) \mathrm{dx}\right) + 2 \setminus \ln 2 \setminus \left(x\right) + C$

$= 2 \left(x \setminus \ln x - x\right) + 2 x \setminus \ln 2 + C$

$= 2 x \setminus \ln \left(2 x\right) - 2 x + C$

Jul 28, 2018

#### Answer:

The answer is $= 2 x \ln 2 x - 2 x + C$

#### Explanation:

A slightly different method

The integral is

$I = \int \ln {\left(2 x\right)}^{2} \mathrm{dx} = 2 \int \ln 2 x \mathrm{dx}$

Perforn an integration by parts

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

$u = \ln 2 x$, $\implies$, $u ' = \frac{1}{2} x \cdot 2 = \frac{1}{x}$

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

Therefore,

$I = 2 x \ln 2 x - 2 \int \frac{1}{x} \cdot x \mathrm{dx}$

$= 2 x \ln 2 x - 2 \int \mathrm{dx}$

$= 2 x \ln 2 x - 2 x + C$