Answer
$$\frac{1}{2}$$
Work Step by Step
$$\int_{1}^{9} \frac{1}{\sqrt{x} (1+\sqrt{x})^2} dx$$
$u=1+\sqrt{x}$
$du=\frac{1}{2} x^{-\frac{1}{2}} dx$
$2du=\frac{1}{\sqrt{x}} dx$
$$2\int_{1}^{9} \frac{1}{u^2} du$$
$2 * _{1}^{9} |-u^{-1}$
$-2 * _{1}^{9} | \frac{1}{1+\sqrt{x}}$
$-2(\frac{1}{4} - \frac{1}{2})$
$$-2(-\frac{1}{4})=\frac{1}{2}$$
