Question

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

Answer 1

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.