Answer
$${\text{The improper integral diverges}}$$
Work Step by Step
$$\eqalign{
& \int_6^\infty {\frac{{d\theta }}{{\sqrt {{\theta ^2} + 1} }}} \cr
& {\text{By the definition of the improper integrals, we have }} \cr
& \int_a^\infty {f\left( x \right)dx = \mathop {\lim }\limits_{b \to \infty } } \int_a^b {f\left( x \right)} dx \cr
& {\text{then}}{\text{,}} \cr
& \int_6^\infty {\frac{{d\theta }}{{\sqrt {{\theta ^2} + 1} }}} = \mathop {\lim }\limits_{b \to \infty } \int_6^b {\frac{{d\theta }}{{\sqrt {{\theta ^2} + 1} }}} \cr
& \cr
& {\text{Integrate by tables using }}\int {\frac{{du}}{{\sqrt {{u^2} + {a^2}} }}} = \ln \left( {u + \sqrt {{u^2} + {a^2}} } \right) + C \cr
& \mathop {\lim }\limits_{b \to \infty } \int_6^b {\frac{{d\theta }}{{\sqrt {{\theta ^2} + 1} }}} = \mathop {\lim }\limits_{b \to \infty } \left[ {\ln \left( {\theta + \sqrt {{\theta ^2} + 1} } \right)} \right]_6^b \cr
& = \mathop {\lim }\limits_{b \to \infty } \left[ {\ln \left( {b + \sqrt {{b^2} + 1} } \right)} \right] - \mathop {\lim }\limits_{b \to \infty } \left[ {\ln \left( {6 + \sqrt {{6^2} + 1} } \right)} \right] \cr
& \cr
& {\text{Evaluate the limit when }}b \to \infty \cr
& = \ln \left( {\infty + \sqrt {{\infty ^2} + 1} } \right) - \ln \left( {6 + \sqrt {{6^2} + 1} } \right) \cr
& = \infty \cr
& \cr
& {\text{Then}}{\text{, the improper integral diverges}} \cr} $$
