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.!

What is the antiderivative of xlnxxlnx?