2. $\displaystyle \int_0^1 {\frac{1} {{1 + \sqrt x }}\,dx} .$
By setting $\displaystyle u = 1 + \sqrt x \implies du = \frac{1} {{2\sqrt x }}\,dx\,\therefore \,du = \frac{1} {{2(u - 1)}}\,dx,$ hence
$\displaystyle 2\int_1^2 {\frac{{u - 1}} {u}\,du} ,$ now this is routine.