Answer
$$\frac{\pi }{{2ab}}$$
Work Step by Step
$$\eqalign{
& \int_0^{ + \infty } {\frac{{dx}}{{{a^2} + {b^2}{x^2}}},\,\,\,\,\,\,\,\,a,b > 0} \cr
& {\text{using the definition 7}}{\text{.8}}{\text{.1 of improper integrals}} \cr
& \,\,\,\int_c^{ + \infty } {f\left( x \right)} dx = \mathop {\lim }\limits_{d \to + \infty } \int_c^d {f\left( x \right)} dx \cr
& \int_0^{ + \infty } {\frac{{dx}}{{{a^2} + {b^2}{x^2}}} = } \mathop {\lim }\limits_{d \to + \infty } \int_0^d {\frac{{dx}}{{{a^2} + {b^2}{x^2}}}} \cr
& = \frac{1}{b}\mathop {\lim }\limits_{d \to + \infty } \int_0^d {\frac{{bdx}}{{{a^2} + {{\left( {bx} \right)}^2}}}} \cr
& \cr
& {\text{Integrate by using the formula }}\int {\frac{{du}}{{{a^2} + {u^2}}}} = \frac{1}{a}{\tan ^{ - 1}}\left( {\frac{u}{a}} \right) + C \cr
& = \frac{1}{b}\mathop {\lim }\limits_{d \to + \infty } \left[ {\frac{1}{a}{{\tan }^{ - 1}}\left( {\frac{{bx}}{a}} \right)} \right]_0^d \cr
& = \frac{1}{{ab}}\mathop {\lim }\limits_{d \to + \infty } \left[ {{{\tan }^{ - 1}}\left( {\frac{{bx}}{a}} \right)} \right]_0^d \cr
& = \frac{1}{{ab}}\mathop {\lim }\limits_{d \to + \infty } \left[ {{{\tan }^{ - 1}}\left( {\frac{{bd}}{a}} \right) - {{\tan }^{ - 1}}\left( {\frac{0}{a}} \right)} \right] \cr
& = \frac{1}{{ab}}\mathop {\lim }\limits_{d \to + \infty } \left[ {{{\tan }^{ - 1}}\left( {\frac{{bd}}{a}} \right)} \right] \cr
& \cr
& {\text{calculate the limit when }}b \to + \infty \cr
& = \frac{1}{{ab}}{\tan ^{ - 1}}\left( \infty \right) \cr
& = \frac{1}{{ab}}\left( {\frac{\pi }{2}} \right) \cr
& {\text{simplify}} \cr
& = \frac{\pi }{{2ab}} \cr
& \cr
& {\text{then}}{\text{,}} \cr
& {\text{The integral converges to }}\frac{\pi }{{2ab}} \cr} $$
