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

3 Answers
Jun 17, 2015

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

Explanation:

I would try using Substitution and By Parts (twice):
enter image source here

Jun 17, 2015

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.

Jun 18, 2015

#= 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.