Answer
$$\frac{1}{{\sqrt 5 }}\left( {\operatorname{arcsec} \frac{{16}}{{\sqrt 5 }} - \operatorname{arcsec} \frac{4}{{\sqrt 5 }}} \right)$$
Work Step by Step
$$\eqalign{
& \int_1^4 {\frac{1}{{x\sqrt {16{x^2} - 5} }}} dx \cr
& {\text{Let }}u = 4x,{\text{ }}x = \frac{u}{4},{\text{ }}dx = \frac{1}{4}du \cr
& {\text{The new limits of integration are}} \cr
& x = 1,{\text{ }}u = 4\left( 1 \right) = 4 \cr
& x = 4,{\text{ }}u = 4\left( 4 \right) = 16 \cr
& {\text{Substituting}} \cr
& \int_1^4 {\frac{1}{{x\sqrt {16{x^2} - 5} }}} dx = \int_4^{16} {\frac{1}{{\frac{u}{4}\sqrt {16{{\left( {\frac{u}{4}} \right)}^2} - 5} }}\left( {\frac{1}{4}} \right)} du \cr
& = \int_4^{16} {\frac{1}{{u\sqrt {{u^2} - 5} }}} du \cr
& {\text{Use the formula }}\int {\frac{{du}}{{u\sqrt {{u^2} - {a^2}} }} = \frac{1}{a}\operatorname{arcsec} \frac{{\left| u \right|}}{a} + C} \cr
& = \left[ {\frac{1}{{\sqrt 5 }}\operatorname{arcsec} \frac{{\left| u \right|}}{{\sqrt 5 }}} \right]_4^{16} \cr
& {\text{Evaluating}} \cr
& = \frac{1}{{\sqrt 5 }}\left[ {\operatorname{arcsec} \frac{{\left| {16} \right|}}{{\sqrt 5 }} - \operatorname{arcsec} \frac{{\left| 4 \right|}}{{\sqrt 5 }}} \right] \cr
& = \frac{1}{{\sqrt 5 }}\left( {\operatorname{arcsec} \frac{{16}}{{\sqrt 5 }} - \operatorname{arcsec} \frac{4}{{\sqrt 5 }}} \right) \cr} $$
