Question

What is `int (lnx) / x^(1/2)dx`?

Answer 1

Put this in your mathematical toolbox: `int x^n lnx dx` can be done by parts. (`u=lnx` and `dv = x^n dx`)

Explanation:

In this case:

`intx^(-1/2) lnx dx`

Let `u = lnx` and `dv = x^(-1/2) dx`.
So that `du = 1/x dx` and `v = 2x^(1/2)`.

`uv-vdu = 2x^(1/2)lnx-2 int x^(1/2) 1/x dx`

`= 2x^(1/2)lnx-2 int x^(-1/2)dx`

`= 2x^(1/2)lnx-4x^(1/2) +C`