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

2 Answers
May 21, 2015

#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#

May 21, 2015

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#