$\displaystyle \int_0^9 \dfrac{1}{\sqrt{1 + \sqrt{x}}} \, dx$
Let
$\sqrt{1 + \sqrt{x}} = u$
$1 + \sqrt{x} = u^2$
$x = (u^2 - 1)^2$
$dx = 2(u^2 - 1)(2u \, du) = 4u(u^2 - 1) \, du$
When x = 0, $u = \sqrt{1 + \sqrt{0}} = 1$
When x = 9, $u = \sqrt{1 + \sqrt{9}} = 2$
Hence,
$\displaystyle \int_0^9 \dfrac{1}{\sqrt{1 + \sqrt{x}}} \, dx = \int_1^2 \dfrac{4u(u^2 - 1) \, du}{u}$
$\displaystyle \int_0^9 \dfrac{1}{\sqrt{1 + \sqrt{x}}} \, dx = \int_1^2 (4u^2 - 4) \, du$
$\displaystyle \int_0^9 \dfrac{1}{\sqrt{1 + \sqrt{x}}} \, dx = \left[ \dfrac{4u^3}{3} - 4u \right]_1^2$
$\displaystyle \int_0^9 \dfrac{1}{\sqrt{1 + \sqrt{x}}} \, dx = \dfrac{4(2^3 - 1^3)}{3} - 4(2 - 1)$
$\displaystyle \int_0^9 \dfrac{1}{\sqrt{1 + \sqrt{x}}} \, dx = 5.333$ ← Answer: [ C ]
