How do you integrate $int x^2lnx$ by integration by parts method?

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Answer 1 · 2017-02-11T01:30:46

Use integration by parts. Let $u = lnx$ and $dv = x^2dx$ Then $du = 1/xdx$ and $v = 1/3x^3$ .

$intx^2lnx dx= 1/3x^3lnx - int(1/3x^3 * 1/x)dx$

$intx^2lnxdx = 1/3x^3lnx - int1/3x^2dx$

$intx^2lnx dx= 1/3x^3lnx - 1/9x^3 + C$

Hopefully this helps!

How do you integrate int x^2lnxint x^2lnx by integration by parts method?