What is the antiderivative of #xln(x) - x#?

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Answer 1 · 2016-06-28T08:26:31

#= x^2 / 2 ln x - 3/4x^2 + C#

#int dx \ (x ln x - x)#

for first term, #x ln x #, we use IBP

#u = ln x, u' = 1/x#
#v' = x, v = x^2/2#

#\implies x^2/2 ln x - int dx \ (x/2)#

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

for the second term

#- int dx \ (x) = - x^2 / 2#

So the integral is

#= x^2 / 2 ln x - x^2 /4 color{red}{- x^2/2 + C}#

#= x^2 / 2 ln x - 3/4x^2 + C#