Question

How do you integrate `ln(2x)`?

Answer 1

`x(ln(2x)-1)+C`

Explanation:

`I=intln(2x)dx`

We should use integration by parts in the absence of all other possible integration strategies. Integration by parts takes the form `intudv=uv-intvdu`. For `intln(2x)dx`, let:

`{(u=ln(2x),=>,du=2/(2x)=1/x),(dv=dx,=>,v=x):}`

Thus:

`I=uv-intvdu=xln(2x)-intx1/xdx`

`I=xln(2x)-intdx`

`I=xln(2x)-x`

`I=x(ln(2x)-1)+C`