Answer
$2\left[ {{{\tan }^{ - 1}}\left( 2 \right) - \frac{\pi }{4}} \right]$
Work Step by Step
$$\eqalign{
& \int_1^4 {\frac{1}{{\left( {x + 1} \right)\sqrt x }}} dx \cr
& {\text{Let }}u = \sqrt x \to du = \frac{1}{{2\sqrt x }}dx,{\text{ }}dx = 2\sqrt x du \cr
& {\text{Finding the new limits of integration}} \cr
& x = 4 \to u = \sqrt 4 = 2 \cr
& x = 1 \to u = \sqrt 1 = 1 \cr
& {\text{Applying the substitution}}{\text{, we obtain}} \cr
& \int_1^4 {\frac{1}{{\left( {x + 1} \right)\sqrt x }}} dx = \int_1^2 {\frac{1}{{\left( {{u^2} + 1} \right)\sqrt x }}\left( {2\sqrt x } \right)} du \cr
& = \int_1^2 {\frac{2}{{{u^2} + 1}}du} \cr
& {\text{Integrating}} \cr
& = 2\left[ {{{\tan }^{ - 1}}u} \right]_1^2 \cr
& = 2\left[ {{{\tan }^{ - 1}}\left( 2 \right) - {{\tan }^{ - 1}}\left( 1 \right)} \right] \cr
& = 2\left[ {{{\tan }^{ - 1}}\left( 2 \right) - \frac{\pi }{4}} \right] \cr} $$
