How do you integrate $int x^4lnx$ by integration by parts method?

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Answer 1 · 2016-09-24T23:55:27

$int x^n lnx dx = 1/(n+1)(x^(n+1)lnx-x^(n+1)/(n+1))+C$

$d/(dx)(x^n lnx) = n x^(n-1)lnx+x^(n-1)$ or

$(n+1)int x^n lnx dx = x^(n+1)lnx-int x^n dx$

so

$int x^n lnx dx = 1/(n+1)(x^(n+1)lnx-x^(n+1)/(n+1))+C$

How do you integrate int x^4lnxint x^4lnx by integration by parts method?