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

1 Answer
Feb 16, 2017

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

Explanation:

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