How do you find the integral of #f(x)=x^2ln2x# using integration by parts?

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Answer 1 · 2018-07-04T08:49:16

#I=x^3/3[ln(2x)-1/3]+c#

Here,

#I=intx^2ln(2x)dx#

#color(blue)(int(u*v)dx=uintvdx-int(u'intvdx)dx#

#Let , u=ln(2x) and v=x^2#

#=>u'=1/(2x)*2=1/x and intvdx=x^3/3#

So,

#I=ln(2x)(x^3/3)-int(1/x xx x^3/3)dx#

#=>I=x^3/3ln(2x)-1/3intx^2dx#

#=>I=x^3/3ln(2x)-1/3 xx x^3/3+c#

#=>I=x^3/3[ln(2x)-1/3]+c#