Question

How do you find the integral `(ln x)^2`?

Answer 1

I found:
`int[ln(x)]^2dx=xln^2(x)-2xln(x)+2x+c`

Explanation:

I would try using Substitution and By Parts (twice):

Answer 2

I get the same answer as Gio,
`int(lnx)^2dx=x(lnx)^2-2xlnx+2x+C`

But the details of my solution are different.

Explanation:

`int(lnx)^2dx`

Use integration by parts:

Let `u = (lnx)^2` and `dv = dx`, so we have

`du = 2/x lnx dx` and `v = x`.

`int(lnx)^2dx = x (lnx)^2 - int x* 2/xlnx dx`

`= x (lnx)^2 -2 int lnx dx`

`= x (lnx)^2 -2 [ xlnx - x]+C`

`= x(lnx)^2-2xlnx+2x+C`

Note
If you don't know `int lnx dx`, use integration by parts with

`u = lnx` and `dv = dx`.

General note
In general to integrate: `int x^r ln x dx` use `u=x^r` and `dv = lnx` -- unless `r = -1` in which case substitution `u = lnx` is easier.

Answer 3

`= intlnxlnxdx`

As funny as it sounds, I'm just going to do it like this. It's how I did it the first time I've seen this one. Let:
`u = lnx`
`du = 1/xdx`
`dv = lnxdx`
`v = xlnx - x`

To do `int lnx`:
Let:
`u = lnx`
`dv = 1dx`
`du = 1/xdx`
`v = x`

`xlnx - intx/xdx = xlnx - x + C`

Anyways, continuing on:

`=> lnx(xlnx - x) - int(xlnx-x)/xdx`

`= xln^2x -xlnx - intlnx-1dx`

`= xln^2x -xlnx - ((xlnx-x)-x)`

`= xln^2x -xlnx - xlnx+x+x + C`

`= xln^2x -2xlnx + 2x + C`

Cool, still got the same answer.