Find the Integral of dx / sqrt(1 + sqrt(x))

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 ]