Answer
$$S = \pi \left( {\sqrt 2 - \frac{{\sqrt {257} }}{{16}} + \ln \left( {\frac{{16 + \sqrt {257} }}{{1 + \sqrt 2 }}} \right)} \right)$$
Work Step by Step
$$\eqalign{
& y = \frac{1}{x},{\text{ 1}} \leqslant x \leqslant 4 \cr
& {\text{The area of surface is given by the formula}} \cr
& S = \int_a^b {2\pi f\left( x \right)} \sqrt {1 + {{\left[ {f'\left( x \right)} \right]}^2}} dx \cr
& {\text{,then}} \cr
& S = 2\pi \int_1^4 {\frac{1}{x}} \sqrt {1 + {{\left( { - \frac{1}{{{x^2}}}} \right)}^2}} dx \cr
& S = 2\pi \int_1^4 {\frac{1}{x}} \sqrt {1 + \frac{1}{{{x^4}}}} dx \cr
& S = 2\pi \int_1^4 {\frac{1}{{{x^3}}}} \sqrt {{x^4} + 1} dx \cr
& {\text{Let }}u = {x^2},{\text{ }}du = 2xdx \cr
& {\text{The new limits of integration are:}} \cr
& x = 4 \Rightarrow u = 16 \cr
& x = 1 \Rightarrow u = 1 \cr
& S = 2\pi \int_1^{16} {\frac{{\sqrt {1 + {u^2}} }}{{2{u^2}}}} du \cr
& S = \pi \int_1^{16} {\frac{{\sqrt {1 + {u^2}} }}{{{u^2}}}} du \cr
& {\text{Integrate by the formula }} \cr
& \int {\frac{{\sqrt {{u^2} + {a^2}} }}{{{u^2}}}du = - \frac{{\sqrt {{u^2} + {a^2}} }}{u} + \ln \left( {u + \sqrt {{u^2} + {a^2}} } \right) + C} \cr
& S = \pi \left[ { - \frac{{\sqrt {1 + {u^2}} }}{u} + \ln \left( {u + \sqrt {1 + {u^2}} } \right)} \right]_1^{16} \cr
& S = \pi \left[ { - \frac{{\sqrt {1 + {{16}^2}} }}{{16}} + \ln \left( {16 + \sqrt {1 + {{16}^2}} } \right)} \right] \cr
& {\text{ }} - \left[ { - \frac{{\sqrt {1 + {{16}^2}} }}{{16}} + \ln \left( {16 + \sqrt {1 + {{16}^2}} } \right)} \right] \cr
& {\text{Simplifying}} \cr
& S = \pi \left( {\sqrt 2 - \frac{{\sqrt {257} }}{{16}} + \ln \left( {16 + \sqrt {257} } \right) + \ln \left( {1 + \sqrt 2 } \right)} \right) \cr
& S = \pi \left( {\sqrt 2 - \frac{{\sqrt {257} }}{{16}} + \ln \left( {\frac{{16 + \sqrt {257} }}{{1 + \sqrt 2 }}} \right)} \right) \cr} $$
