How do you integrate #int u^5lnu# by integration by parts method?

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Answer 1 · 2016-09-05T13:09:20

# = u^6/6 ( lnu - 1/6) + C#

tactically, the #ln u# is the bit you want to differentiate cos that just turns it into another #u# term. So we approach the IBP as

#int u^5lnu \ du #

# = int (u^6/6)' \ lnu \ du #

and so by IBP we get to differentiate it in the next line

# = u^6/6 \ lnu - int u^6/6 \ ( lnu)' \ du #

# = u^6/6 \ lnu - int u^6/6 \ ( 1/u) \ du # !!

# = u^6/6 \ lnu - int u^5/6 \ du #

# = u^6/6 \ lnu - u^6/36 + C#

# = u^6/6 ( lnu - 1/6) + C#