How do you evaluate the integral #int xlnxdx#?

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Answer 1 · 2017-01-10T11:11:14

#1/4x^2(2lnx-1)+C#

This needs the method of 'integration by parts'. The formula of which needs to be known.

#intu(dv)/(dx)dx=uv-intv(du)/(dx)dx#

Care must be taken in the choice of #u# & #(dv)/(dx)# otherwise you end up with a more difficult, or impossible integration!

When logs are involved take the #u# as the log part, the reason will become clear as we work through this problem.

#int xlnxdx#

#u=lnx=>(du)/(dx)=1/x#

#(dv)/(dx)=x=>v=1/2x^2#

#I=intu(dv)/(dx)dx=uv-intv(du)/(dx)dx#

#I=1/2x^2lnx-int1/2x^2xx 1/xdx#

#I=1/2x^2lnx-1/2intxdx#

#I=1/2x^2lnx-1/4x^2 +C#

#I=1/4x^2(2lnx-1)+C#