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

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Answer 1 · 2016-06-21T09:49:57

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

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

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

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

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

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