Answer
$$ - \frac{{{{\left( {25{x^2} + 4} \right)}^{3/2}}}}{{12{x^3}}} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{\sqrt {25{x^2} + 4} }}{{{x^4}}}} dx \cr
& {\text{Rewrite}} \cr
& \int {\frac{{\sqrt {25{x^2} + 4} }}{{{x^4}}}} dx = \int {\frac{{\sqrt {{{\left( {5x} \right)}^2} + {{\left( 2 \right)}^2}} }}{{{x^4}}}} dx \cr
& {\text{Let }}u = 5x,{\text{ and }}du = 5dx,{\text{ }}dx = \frac{1}{5}du \cr
& \int {\frac{{\sqrt {{{\left( {5x} \right)}^2} + {{\left( 2 \right)}^2}} }}{{{x^4}}}} dx = \int {\frac{{\sqrt {{u^2} + {2^2}} }}{{{{\left( {u/5} \right)}^4}}}} \left( {\frac{1}{5}} \right)du \cr
& = 125\int {\frac{{\sqrt {{u^2} + {2^2}} }}{{{u^4}}}} du \cr
& {\text{Refer to the triangle below}} \cr
& \tan \theta = \frac{u}{2},{\text{ }}u = 2\tan \theta ,{\text{ }}du = 2{\sec ^2}\theta d\theta \cr
& 125\int {\frac{{\sqrt {{u^2} + {2^2}} }}{{{u^4}}}} du = 125\int {\frac{{\sqrt {{{\left( {2\tan \theta } \right)}^2} + {2^2}} }}{{{{\left( {2\tan \theta } \right)}^4}}}} \left( {2{{\sec }^2}\theta } \right)d\theta \cr
& = \frac{{125}}{8}\int {\frac{{\sqrt {4{{\tan }^2}\theta + 4} }}{{{{\tan }^4}\theta }}} {\sec ^2}\theta d\theta \cr
& = \frac{{125}}{8}\int {\frac{{2\sqrt {{{\tan }^2}\theta + 1} }}{{{{\tan }^4}\theta }}} {\sec ^2}\theta d\theta \cr
& = \frac{{125}}{4}\int {\frac{{\sec \theta }}{{{{\tan }^4}\theta }}} {\sec ^2}\theta d\theta \cr
& = \frac{{125}}{4}\int {\frac{{{{\sec }^3}\theta }}{{{{\tan }^4}\theta }}} d\theta \cr
& = \frac{{125}}{4}\int {\left( {\frac{1}{{{{\cos }^3}\theta }}} \right)\left( {\frac{{{{\cos }^4}\theta }}{{{{\sin }^4}\theta }}} \right)} d\theta \cr
& = \frac{{125}}{4}\int {\frac{{\cos \theta }}{{{{\sin }^4}\theta }}} d\theta \cr
& {\text{Integrating}} \cr
& = \frac{{125}}{4}\left( { - \frac{1}{{3{{\sin }^3}\theta }}} \right) + C \cr
& = - \frac{{125}}{{12}}{\csc ^3}\theta + C \cr
& {\text{Refer to the triangle}} \cr
& \csc \theta = \frac{{\sqrt {{u^2} + {2^2}} }}{u} \cr
& = - \frac{{125}}{{12}}{\left( {\frac{{\sqrt {{u^2} + {2^2}} }}{u}} \right)^3} + C \cr
& = - \frac{{125}}{{12}}\left( {\frac{{{{\left( {{u^2} + {2^2}} \right)}^{3/2}}}}{{{u^3}}}} \right) + C \cr
& = - \frac{{125{{\left( {{u^2} + {2^2}} \right)}^{3/2}}}}{{12{u^3}}} + C \cr
& {\text{Substitute back }}u = 5x \cr
& = - \frac{{125{{\left( {25{x^2} + 4} \right)}^{3/2}}}}{{12{{\left( {5x} \right)}^3}}} + C \cr
& = - \frac{{{{\left( {25{x^2} + 4} \right)}^{3/2}}}}{{12{x^3}}} + C \cr} $$
