int(ln(3x))^2 dx=x(ln(3x))^2-2xln(3x)+2x+C∫(ln(3x))2dx=x(ln(3x))2−2xln(3x)+2x+C
We try u=3xu=3x or u=lnvu=lnv , with v=3xv=3x, but neither of those work. So we give up. No, not seriously,
Substitution doesn't seem to be working, so try the next less simple method Integration by Parts.
Try something, judge whether we seem to be making progress.
(Yes, really. There's no cookbook to solve every problem.)
Well, we can't integrate, so let's differentiate.
Let u=(ln(3x))^2u=(ln(3x))2 and dv=dxdv=dx.
With these choices, we get
du=2(ln(3x))*1/xdxdu=2(ln(3x))⋅1xdx and v=xv=x.
Applying Integration by Parts gives us
int(ln(3x))^2 dx=uv-intvdu∫(ln(3x))2dx=uv−∫vdu
=x(ln(3x))^2-intx*2(ln(3x))*1/x dx=x(ln(3x))2−∫x⋅2(ln(3x))⋅1xdx (looks worse, but)
=x(ln(3x))^2-2int(ln(3x)) dx=x(ln(3x))2−2∫(ln(3x))dx. So all we need to do is remember or figure out how to integrate lnu lnu dudu.
(By parts, to get intlnu du = u ln u-u+C∫lnudu=ulnu−u+C)
With u=3xu=3x, we can finish the integral to get:
int(ln(3x))^2 dx=x(ln(3x))^2-2(xln(3x)+x)+C∫(ln(3x))2dx=x(ln(3x))2−2(xln(3x)+x)+C
int(ln(3x))^2 dx=x(ln(3x))^2-2xln(3x)+2x+C∫(ln(3x))2dx=x(ln(3x))2−2xln(3x)+2x+C
Note: you can learn a lot, by checking the answer by differentiating.