Calculus: Early Transcendentals, Chapter 7, 7.1, Section 7.1, Problem 15

int (ln x)^2dx
To evaluate, apply integration by parts int udv= uv - vdu .
So let
u = (lnx)^2
and
dv = dx
Then, differentiate u and integrate dv.
du = 2lnx * 1/x dx = (2lnx)/x dx
and
v = intdx = x
Plug-in them to the formula. So the integral becomes:
int (ln x)^2dx
= (lnx)^2 *x - int x * (2lnx)/x dx
= x(lnx)^2 -2int lnx dx
To take integral of ln x, apply integration by parts again.
So let
u_2 = ln x
and
dv_2 = dx
Then, differentiate u and integrate dv.
du_2 = 1/x dx
and
v_2= int dx = x
So the integral becomes:
=x(lnx)^2 - 2( lnx * x - int x * 1/x dx)
= x(lnx)^2 - 2(xlnx - int dx)
=x(lnx)^2 - 2(xlnx - x)
=x(lnx)^2 - 2xlnx + 2x

Therefore, int (lnx)^2dx = x(lnx)^2 - 2xlnx + 2x .