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

1 Answer
Jul 31, 2014

Our objective is to reduce the power of lnx so that the integral is easier to evaluate.

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

udv=uvvdu

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

Therefore,

du=2lnxxdx

and

v=x.

Now, assembling the pieces together, we get:

(lnx)2dx=x(lnx)22xlnxxdx

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

(lnx)2dx=x(lnx)22lnxdx

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

Thus, du=1xdx and v=x.

Assembling gives us:

(lnx)2dx=x(lnx)22(xlnxxxdx)

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

(lnx)2dx=x(lnx)22xlnx+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 (lnx)2 down to lnx, and then down to 1x. In the end, some x's canceled off, and it became easier to integrate.