# How do you integrate int( 3x ln(2x) dx}?

Apr 10, 2018

$I = \frac{3 {x}^{2}}{4} \left(\ln \left(4 {x}^{2}\right) - 1\right) + C$

#### Explanation:

Here,

$I = \int \left(3 x\right) \ln \left(2 x\right) \mathrm{dx}$

$\text{Using "color(blue)"Integration by Parts}$

color(red)(int(u*v)dx=uintvdx-int((du)/(dx)intvdx)dx

Let, $u = \ln \left(2 x\right) \mathmr{and} v = 3 x$,we get

$\frac{\mathrm{du}}{\mathrm{dx}} = \frac{1}{2 x} \cdot 2 = \frac{1}{x} \mathmr{and} \int v \mathrm{dx} = \frac{3 {x}^{2}}{2}$

$\implies I = \ln \left(2 x\right) \frac{3 {x}^{2}}{2} - \int \left(\frac{1}{x} \frac{3 {x}^{2}}{2}\right) \mathrm{dx}$

$= \frac{3 {x}^{2}}{2} \ln \left(2 x\right) - \frac{3}{2} \int x \mathrm{dx} + c$

$= \frac{3 {x}^{2}}{2} \ln \left(2 x\right) - \frac{3}{2} \cdot {x}^{2} / 2 + C$

$= \frac{3 {x}^{2}}{4} \left(2 \ln \left(2 x\right) - 1\right) + C$

$= \frac{3 {x}^{2}}{4} \left(\ln {\left(2 x\right)}^{2} - 1\right) + C$

$= \frac{3 {x}^{2}}{4} \left(\ln {\left(2 x\right)}^{2} - 1\right) + C$

$= \frac{3 {x}^{2}}{4} \left(\ln \left(4 {x}^{2}\right) - 1\right) + C$