The first thing to do is see about getting rid of that square root. Since x has to be positive anyway I would suggest the substitution $\displaystyle x = y^2$. Then $\displaystyle dx = 2y~dy$ and your integral now becomes:
$\displaystyle \int_0^4 \dfrac{1}{1 + \sqrt{x}}~dx = \int_0^2 \dfrac{1}{1 + y}~2y~dy$
which is a much more familiar form. Can you take it from here?
-Dan
Here's another way:
Let $\displaystyle \ u = 1 + \sqrt{x}$
As the limits of integration for x are from 0 to 4, then for u, they are from 1 to 3.
$\displaystyle u = 1 + \sqrt{x} $
$\displaystyle u - 1 = \sqrt{x} $
$\displaystyle (u - 1)^2 = x $
$\displaystyle dx = 2(u - 1)du$
The integrand will be $\displaystyle \ \ \dfrac{2(u - 1)du}{u} \ = $
$\displaystyle 2\bigg(1 - \dfrac{1}{u}\bigg)du$.
And then you can continue.