# How do you integrate int x^3 ln 2x dx  using integration by parts?

Jun 21, 2016

$\ln \left(2 x\right) \setminus \frac{{x}^{4}}{4} - \setminus \frac{{x}^{4}}{16} + C$

#### Explanation:

Using the formulation $\setminus \int u v ' \setminus \mathrm{dx} = \left[u v\right] - \setminus \int u ' v \setminus \mathrm{dx}$

with $u = \setminus \ln \left(2 x\right) , u ' = \setminus \frac{1}{2 x} .2 = \setminus \frac{1}{x}$

and $v ' = {x}^{3} , v = \setminus \frac{{x}^{4}}{4}$, we get

$\ln \left(2 x\right) \setminus \frac{{x}^{4}}{4} - \setminus \int \setminus \frac{1}{x} . \setminus \frac{{x}^{4}}{4} \setminus \mathrm{dx}$

$= \ln \left(2 x\right) \setminus \frac{{x}^{4}}{4} - \setminus \int \setminus \frac{{x}^{3}}{4} \setminus \mathrm{dx}$

$= \ln \left(2 x\right) \setminus \frac{{x}^{4}}{4} - \setminus \frac{{x}^{4}}{16} + C$