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

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Answer 1 · 2016-07-12T19:25:44

reading this as : $int \ ln(2x^2) \ dx$

then

$= x ( ln2x^2 - 2) + C$

$int \ ln(2x^2) \ dx$

using IBP

$= int \d/dx(x) * ln2x^2 \ dx$

$= x * ln2x^2 - int \ x * d/dx ( ln2x^2) \ dx$

$= x ln2x^2 - int \ x * 1/(2x^2) 4x \ dx$

$= x ln2x^2 - 2 int \ dx$

$= x ln2x^2 - 2x + C$

$= x ( ln2x^2 - 2) + C$

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