What is the antiderivative of #ln(x)^2#?

1 Answer
Oct 25, 2016

Answer:

# int (lnx)^2dx = x(lnx)^2 +2x +C#

Explanation:

You should learn the IBP formula:
# int u(dv)/dxdx=uv - int v (du)/dxdx #

So essentially we are looking for one function that simplifies when it is differentiated, and one that simplifies when integrated (or at least is integrable).

We also need to know that #int lnxdx=xlnx-x# (either learn or use IBP):

Let # {(u=lnx, => ,(du)/dx=1/x),((dv)/dx=lnx,=>,v =xlnx-x ):}#

Then IBP gives;
# int lnx lnx dx=lnx(xlnx-x) - int (xlnx-x)1/xdx #
# :. int (lnx)^2 dx = x(lnx)^2 - xlnx - int (lnx-1)dx #
# :. int (lnx)^2 dx = x(lnx)^2 - xlnx - (xlnx-x-x) #
# :. int (lnx)^2 dx = x(lnx)^2 - xlnx + xlnx +2x #

And so we have:
# :. int (lnx)^2dx = x(lnx)^2 +2x +C#