Question

How do I evaluate `int(ln(3x))^2 dx`?

Answer 1

Using Integration by substitution I set: `ln(3x)=t` so:
`3x=e^t`
`x=e^t/3` and
`dx=1/3e^tdt`

Your integral becomes:

`intt^2*1/3e^tdt=`

Which can now be solved by parts (twice).

By parts you have:

`intf(x)*g(x)dx=F(x)*g(x)-intF(x)*g'(x)dx`

Where:

`F(x)=intf(x)dy`
`g'(x)` is the derivative of `g(x)`
We can choose:
`f(x)=e^t`
`g(x)=t^2`

The integral becomes:

`intt^2*1/3e^tdt=1/3[e^t*t^2-inte^t2tdt]=` by parts again:
`=1/3[e^t*t^2-e^t*2t+inte^t*2dt]=`
`1/3[e^t*t^2-e^t*2t+2*e^t]=`
`1/3e^t(t^2-2t+2)`
but `ln(3x)=t` so going back to `x`:
`=1/3*3x(ln^2(3x)-2ln(3x)+2)+c`
`=x(ln^2(3x)-2ln(3x)+2)+c`