Answer
$$\frac{1}{{\sqrt {17} }}{\tan ^{ - 1}}\left( {\frac{x}{{\sqrt {17} }}} \right)$$
Work Step by Step
$$\eqalign{
& \int {\frac{{dx}}{{17 + {x^2}}}} \cr
& {\text{write }}17 + {x^2}{\text{ as }}{\left( {\sqrt {17} } \right)^2} + {\left( x \right)^2} \cr
& \int {\frac{{dx}}{{{{\left( {\sqrt {17} } \right)}^2} + {{\left( x \right)}^2}}}} \cr
& {\text{intgrate by using the formula }}\int {\frac{{du}}{{{a^2} + {u^2}}} = \frac{1}{a}{{\tan }^{ - 1}}\left( {\frac{u}{a}} \right) + C\,\,\,\left( {{\text{see page 419}}} \right)} \cr
& {\text{with }}a = \sqrt {17} {\text{ and }}u = x,\,\,\,\,du = dx \cr
& = \int {\frac{{dx}}{{{{\left( {\sqrt {17} } \right)}^2} + {{\left( x \right)}^2}}}} = \frac{1}{{\sqrt {17} }}{\tan ^{ - 1}}\left( {\frac{x}{{\sqrt {17} }}} \right) \cr} $$
