How do you find the integral of #f(x)=x^5lnx# using integration by parts?

1 Answer
Oct 31, 2015

#I = x^6/6(lnx - 1/6) +C#

Explanation:

#lnx=u => dx/x=du#

#x^5dx=dv => v=x^6/6#

#int udv = uv - int vdu#

#I = int x^5lnxdx = x^6/6lnx - int x^6/6dx/x = x^6/6lnx - 1/6int x^5dx#

#I = x^6/6lnx - 1/6x^6/6 +C#

#I = x^6/6(lnx - 1/6) +C#