Question

Integrate ln(3x)^2dx

Answer 1

$\ln(3x)^2=2\ln(3x)$

Take $y=3x$, so that $\dfrac{\mathrm dy}3=\mathrm dx$, and you have

$\displaystyle\int\ln(3x)^2\,\mathrm dx=\frac23\int\ln y\,\mathrm dy$

Integrate by parts, setting $u=\ln y$ and $\mathrm dv=\mathrm dy$, which give $\mathrm du=\dfrac{\mathrm dy}y$ and $v=y$. Then

$\displaystyle\frac23\int\ln y\,\mathrm dy=\frac23\left(y\ln y-\int\frac yy\,\mathrm dy\right)=\frac23y\ln y-\frac23y+C$

which in terms of $x$ is

$\displaystyle\frac23y\ln (3x)-\frac23(3x)+C$
$\displaystyle\frac23y\ln x-2x+C$

where the last term uses the property that $\ln(ab)=\ln a+\ln b$, and the $\dfrac23\ln3$ term is absorbed into $C$.