Question

Integrate ln(3x)^2dx

Answer 1

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

Take
`y=3x`, so that
`\frac{\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=\frac{\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
`\frac23\ln3` term is absorbed into
`C`.