ind the area under the curve Y = 1 / 1 + (X)^0.5 between x = 0 and x = 1 using the substitution u = 1 + (x)^0.5
$\displaystyle \int_0^1 {\frac{1}
{{1 + \sqrt x }}\,dx} .$
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.