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

1 Answer
Eddie
Jun 21, 2016

# 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#

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