Question

How do you evaluate the integral of `(ln x)^2 dx`?

Answer 1

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

To integrate `(lnx)^2`, let `x= e^y` so that `dx= e^y dy`

`int (lnx)^2 dx= int y^2 e^ydy`. Now integrate by parts,

`y^2 e^y -int 2ye^y dy`. Now again integrate by parts,

`y^2 e^y -2[ ye^y- int e^ydy]`

`y^2e^y -2ye^y +2e^y` +C

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

Answer 2

bp has one great solution Method 1. There are other solutions:

Both of the solution presented below use Integration by Parts.
I use the form:

`int u dv = uv-intvdu`.

Both of the solution presented below use `int lnx dx = xlnx - x +C`, which can be done by integration by parts. (And, of course, verified by differentiating the answer.)

Method 2

`int (lnx)^2 dx`

Let `u = (lnx)^2` and `dv = dx`.

Then `du = (2lnx)/x dx` and `v = x`

Integration by parts gives us:

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

`color(white)"sssssss"` `=x(lnx)^2-2(xlnx - x) +C`

`color(white)"sssssss"` `=x(lnx)^2-2xlnx + 2x +C`

Method 3

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

Let `u=lnx` and `dv = lnx dx`

So, `du = 1/x dx` and `v= xlnx -x`

The parts formula gives us:

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

`color(white)"sssssss"` `=x(lnx)^2-xlnx -int (color(red)(lnx) - color(green)(1))dx`

`color(white)"sssssss"` `=x(lnx)^2-xlnx -(color(red)(xlnx-x) - color(green)(x)) +C`

`color(white)"sssssss"` `=x(lnx)^2-2xlnx +2x +C`