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

1 Answer

Jul 20, 2016

#int x^n log_e x dx = x^{n+1}(((n+1)log_e x-1)/(n+1)^2)#

Explanation:

#d/(dx)(x^{n+1}log_e x) = (n+1)x^nlog_ex+x^n#

then

#(n+1)int x^n log_e x dx = x^{n+1}log_e x - 1/(n+1)x^{n+1}#

so

#int x^n log_e x dx = x^{n+1}(((n+1)log_e x-1)/(n+1)^2)#

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