How do I find the integral #intx^5*ln(x)dx# ?

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Answer 1 · 2014-09-24T04:19:36

By Integration by Parts,

#int x^5lnx dx=x^6/36(6lnx-1)+C#

Let us look at some details.

Let #u=lnx# and #dv=x^5dx#.
#Rightarrow du={dx}/x# and #v=x^6/6#

By Integration by Parts

#int udv=uv-int vdu#,

we have

#int (lnx)cdot x^5dx=(lnx)cdot x^6/6-int x^6/6cdot dx/x#

by simplifying a bit,

#=x^6/6lnx-int x^5/6dx#

by Power Rule,

#=x^6/6lnx-x^6/36+C#

by factoring out #x^6/36#,

#=x^6/36(6lnx-1)+C#