How do I find the integral #int(ln(x))^2dx# ?

1 Answer
Jul 31, 2014

Our objective is to reduce the power of #ln x# so that the integral is easier to evaluate.

We can accomplish this by using integration by parts. Keep in mind the IBP formula:

#int u dv = uv - int v du#

Now, we will let #u = (lnx)^2#, and #dv = dx#.

Therefore,

#du = (2lnx)/x dx#

and

#v = x#.

Now, assembling the pieces together, we get:

#int (ln x)^2 dx = x(ln x)^2 - int (2xlnx)/x dx#

This new integral looks a lot better! Simplifying a bit, and bringing the constant out front, yields:

#int (ln x)^2 dx = x(ln x)^2 - 2 int lnx dx#

Now, to get rid of this next integral, we will do a second integration by parts, letting #u = ln x# and #dv = dx#.

Thus, #du = 1/x dx# and #v = x#.

Assembling gives us:

#int (ln x)^2 dx = x(ln x)^2 - 2(xlnx - int x/x dx)#

Now, all that's left to do is simplify, keeping in mind to add the constant of integration:

#int (ln x)^2 dx = x(ln x)^2 - 2xlnx + 2x + C#

And there we have it. Remember, integration by parts is all about picking #u# so that messy things get eliminated from the integrand. In this case we brought #(ln x)^2# down to #ln x#, and then down to #1/x#. In the end, some #x#'s canceled off, and it became easier to integrate.