What is the antiderivative of #xlnx#?

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Answer 1 · 2016-09-21T07:00:42

#intxlnxdx=x^2/4(2lnx-1)+C, or, x^2/4ln(x^2/e)+C, or ln(x^2/e)^(x^2/4)+C.#

Let #I=intxlnxdx#

We use the following Rule of Integration by Parts (IBP) :

#intuvdx=uintvdx-int[(du)/dxintvdx]dx.# We take,

#u=lnx rArr (du)/dx=1/x, and, v=x rArr intvdx=intxdx=x^2/2#.

Hence,

#I=x^2/2lnx-int[1/x*x^2/2]dx#

#=x^2/2lnx-1/2intxdx#

#=x^2/2lnx-1/2(x^2/2)#

#=x^2/2lnx-x^2/4#

#=x^2/4(2lnx-1),# or,

#=x^2/4(lnx^2-lne)#

#=x^2/4ln(x^2/e)#, or,

#=ln(x^2/e)^(x^2/4).#

Enjoy Maths.!