Question

What is `int ln(2x)`?

Answer 1

I found: `x(ln(2x)-1)`

Explanation:

You can try setting:
`ln(2x)=t` so:
`2x=e^t`
`x=1/2e^t`
`dx=1/2e^tdt`
The integral becomes:
`intln(2x)dx=intte^t1/2dt=`
This can be solved using Integration by Parts:
`=1/2[te^t-inte^t*1dt]=`
`=1/2[te^t-e^t]=`
`=1/2e^t(t-1)`
Back to `x`:
`=(1/2)2x(ln(2x)-1)=x(ln(2x)-1)`

Answer 2

You can go like this

`int ln2xdx=int dx/dx*ln2xdx=xln2x-int x*dln(2x)/dxdx=xln2x-int x*2/(2x)dx=x*ln2x-x+c`

(*) We used integration by parts hence

`int f'(x)*g(x)dx=f(x)*g(x)-int f(x)*g'(x)dx`

where `f(x)=x` and `g(x)=ln2x`