Answer
$$ - 2 - \ln 4$$
Work Step by Step
$$\eqalign{
& \int_4^9 {\frac{{dx}}{{1 - \sqrt x }}} \cr
& {\text{Multiply the integrand by }}\frac{{2\sqrt x }}{{2\sqrt x }} \cr
& = \int {\frac{1}{{1 - \sqrt x }}\left( {\frac{{2\sqrt x }}{{2\sqrt x }}} \right)} dx \cr
& = \int {\frac{{2\sqrt x }}{{2\sqrt x \left( {1 - \sqrt x } \right)}}} dx \cr
& \cr
& {\text{Let }}u = \sqrt x ,\,\,\,du = \frac{1}{{2\sqrt x }}dx \cr
& {\text{Use the substitution}} \cr
& = \int {\frac{{2u}}{{1 - u}}} du \cr
& = - 2\int {\frac{u}{{u - 1}}} du \cr
& = - 2\int {\left( {1 + \frac{1}{{u - 1}}} \right)} du \cr
& = - 2\int {du} - 2\int {\frac{1}{{u - 1}}} du \cr
& \cr
& {\text{Integrating}} \cr
& {\text{ = }} - 2u - 2\ln \left| {u - 1} \right| + C \cr
& {\text{substitute }}u = \sqrt x \cr
& {\text{ = }} - 2\sqrt x - 2\ln \left| {\sqrt x - 1} \right| + C \cr
& {\text{,then}} \cr
& \int_4^9 {\frac{{dx}}{{1 - \sqrt x }}} = \left[ { - 2\sqrt x - 2\ln \left| {\sqrt x - 1} \right|} \right]_4^9 \cr
& = \left( { - 2\sqrt 9 - 2\ln \left| {\sqrt 9 - 1} \right|} \right) - \left( { - 2\sqrt 4 - 2\ln \left| {\sqrt 4 - 1} \right|} \right) \cr
& = \left( { - 6 - 2\ln \left| 2 \right|} \right) - \left( { - 4 - 2\ln \left| 1 \right|} \right) \cr
& = - 6 - \ln 4 + 4 \cr
& = - 2 - \ln 4 \cr} $$