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

Question

Isabella Ackerman · Accepted Answer

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

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

Elijah Abram · Answer

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

But the details of my solution are different.

#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.

Cameron Alexander · Answer

#= 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.

Paisley Johnson · Answer

undefined To find the integral of (ln x)^2, you can use integration by parts. Let u = ln(x) and dv = ln(x) dx. Then differentiate u to get du and integrate dv to get v. After that, apply the integration by parts formula:

∫(ln(x))^2 dx = u*v - ∫v du

Substitute the values of u, v, du, and dv into the formula and perform the necessary calculations to find the integral.