How do you integrate #int x^3lnabsx# from 1 to 2 by integration by parts method?

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Answer 1 · 2016-08-24T10:14:39

#= 4 ln 2 - 15/16 #

#int_1^2 x^3lnabsx \ dx#

#= int_1^2 x^3 ln x \ dx#, as #x >0#

#= int_1^2 d/dx(x^4/4) ln x \ dx#

#= [ x^4/4 ln x ]_1^2 - int_1^2 x^4/4 d/dx( ln x )\ dx#

#= [ x^4/4 ln x ]_1^2 - int_1^2 x^3/4 \ dx#

#= [ x^4/4 ln x - x^4/16 ]_1^2#

#= [ 16/4 ln 2 - 16/16 ] - [ 1/4 (0) - 1/16 ]#

#= 4 ln 2 - 15/16 #