How do you evaluate the integral #int (lnx)^3dx#?

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Answer 1 · 2017-02-16T09:57:08

#int (lnx)^3dx = x((lnx)^3-3(lnx)^2+6lnx-6)+C#

Substitute:

#t= lnx#

#x = e^t#

#dx = e^tdt#

so:

#int (lnx)^3dx = int t^3e^tdt#

Integrate now by parts:

#int t^3e^tdt = int t^3d(e^t) = t^3e^t - 3int t^2e^tdt#

#int t^2e^tdt = int t^2d(e^t) = t^2e^t - 2int te^tdt#

#int te^tdt = int td(e^t) = te^t - int e^tdt = te^t -e^t +C#

Putting it all together:

#int t^3e^tdt = t^3e^t -3 t^2e^t +6 te^t -6e^t +C = e^t(t^3-3t^2+6t-6)+C#

and undoing the substitution:

#int (lnx)^3dx = x((lnx)^3-3(lnx)^2+6lnx-6)+C#