Answer
$$2\left( {4y - 3} \right)$$
Work Step by Step
$$\eqalign{
& \int_1^4 {\frac{{4y - 3}}{{\sqrt x }}} dx \cr
& {\text{the notation }}dx{\text{ indicates integration with respect to }}x,{\text{ }} \cr
& {\text{so we treat }}x{\text{ as variable and }}y{\text{ as a constant}}{\text{. then}} \cr
& \int_1^4 {\frac{{4y - 3}}{{\sqrt x }}} dx = \left( {4y - 3} \right)\int_1^4 {\frac{1}{{\sqrt x }}} dx \cr
& = \left( {4y - 3} \right)\int_1^4 {{x^{ - 1/2}}} dx \cr
& {\text{using the power rule }} \cr
& = \left( {4y - 3} \right)\left( {\frac{{{x^{1/2}}}}{{1/2}}} \right)_1^4 \cr
& = \left( {4y - 3} \right)\left( {2\sqrt x } \right)_1^4 \cr
& {\text{evaluating the limits in the variable }}x \cr
& = \left( {4y - 3} \right)\left( {2\sqrt 4 - 2\sqrt 1 } \right) \cr
& {\text{simplifying}} \cr
& = 2\left( {4y - 3} \right) \cr} $$
