+118703 Hint:
the integral is lnx = the integral of (1*lnx)
do that first.
You need to use integration by parts twice.
I have not done it, I don't have time at present, but that is how I would proceed.
+118703 \displaystyle \int [ln(x)]^2 dx
∫[ln(x)]2dx=∫[ln(x)][ln(x)]dxu=lnxv′=lnxu′=1xv=? −−−−−−−−−−−−−−−−−−−−−∫[ln(x)]dx=∫1∗[ln(x)]dxp=lnxq′=1p′=1xq=xso∫1∗[ln(x)]dx=xlnx−∫1x∗xdx=xlnx−∫1dx=xlnx−x So v=xlnx-x −−−−−−−−−−−−−−−−
Going back to the beginning
∫[ln(x)]2dx=∫[ln(x)][ln(x)]dxu=lnxv′=lnxu′=1xv=xlnx−x =uv−∫vu′dx+c
And you can finish it
LaTex:
\displaystyle \int [ln(x)]^2 dx\\
=\displaystyle \int [ln(x)] [ln(x)] dx\\
u=lnx\qquad v'=lnx\\
u'=\frac{1}{x}\qquad v=?\\~\\
---------------------\\
\displaystyle \int [ln(x)] dx\\
=\displaystyle \int 1*[ln(x)] dx\\
p=lnx\qquad q'=1\\
p'=\frac{1}{x}\qquad q=x\\
so\\
\displaystyle \int 1*[ln(x)] dx\\
=xlnx-\int \frac{1}{x}*x\;dx\\
=xlnx-\int 1\;dx\\
=xlnx-x\\~\\
\text {So v=xlnx-x}\\~\\
----------------\\
