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#