Answer
$$ - \frac{{\sqrt {{x^2} + 25} }}{{25x}} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{dx}}{{{x^2}\sqrt {{x^2} + 25} }}} \cr
& {\text{substitute }}x = 5\tan \theta ,{\text{ }}dx = 5{\sec ^2}\theta d\theta \cr
& = \int {\frac{{5{{\sec }^2}\theta d\theta }}{{25{{\tan }^2}\theta \sqrt {{{\left( {5\tan \theta } \right)}^2} + 25} }}} . \cr
& = \int {\frac{{5{{\sec }^2}\theta d\theta }}{{25{{\tan }^2}\theta \left( 5 \right)\sqrt {{{\tan }^2}\theta + 1} }}} \cr
& {\text{use identity }}{\sec ^2}\theta - 1 = {\tan ^2}\theta \cr
& = \int {\frac{{5{{\sec }^2}\theta d\theta }}{{25{{\tan }^2}\theta \left( 5 \right)\sqrt {{{\sec }^2}\theta } }}} \cr
& = \int {\frac{{{{\sec }^2}\theta d\theta }}{{25{{\tan }^2}\theta \sec \theta }}} \cr
& = \frac{1}{{25}}\int {\frac{{\sec \theta }}{{{{\tan }^2}\theta }}} d\theta \cr
& = \frac{1}{{25}}\int {\frac{{\cos \theta }}{{{{\sin }^2}\theta }}d\theta } \cr
& {\text{simplify by using trig identities}} \cr
& = \frac{1}{{25}}\int {\cot \theta \csc \theta } d\theta \cr
& {\text{find antiderivative}} \cr
& = - \frac{1}{{25}}\csc \theta + C \cr
& {\text{write in terms of }}x{\text{ }} \cr
& = - \frac{{\sqrt {{x^2} + 25} }}{{25x}} + C \cr} $$
