Question

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

Answer 1

`ln(2x) \frac{x^4}{4} - \frac{x^4}{16} + C`

Explanation:

Using the formulation `\int u v' \dx= [uv] - \int u' v \dx`

with `u = \ln (2x), u' = \frac{1}{2x}.2 = \frac{1}{x}`

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

`ln(2x) \frac{x^4}{4} - \int \frac{1}{x} .\frac{x^4}{4} \dx`

`= ln(2x) \frac{x^4}{4} - \int \frac{x^3}{4} \dx`

`= ln(2x) \frac{x^4}{4} - \frac{x^4}{16} + C`