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

1 Answer

Sep 24, 2016

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

Explanation:

#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#

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