Question

What is `int ln(lnx)/x dx`?

Answer 1

`lnxln(lnx)-lnx+C`

Explanation:

`lnx=t => 1/xdx=dt`

`int (ln(lnx))/xdx=int lntdt`

`lnt=u => du=dt/t`

`dt=dv => v=t`

`I=intlntdt = tlnt - inttdt/t=tlnt-intdt = tlnt-t+C`

`I=lnxln(lnx)-lnx+C`