Answer
$$\ln 2$$
Work Step by Step
$$\eqalign{
& \int_3^{ + \infty } {\frac{2}{{{x^2} - 1}}} dx \cr
& {\text{using the definition 7}}{\text{.8}}{\text{.1 of improper integrals}} \cr
& \,\,\,\int_a^{ + \infty } {f\left( x \right)} dx = \mathop {\lim }\limits_{b \to + \infty } \int_a^b {f\left( x \right)} dx \cr
& \cr
& {\text{then}} \cr
& \,\,\,\int_3^{ + \infty } {\frac{2}{{{x^2} - 1}}} dx = \mathop {\lim }\limits_{b \to + \infty } \int_3^b {\frac{2}{{{x^2} - 1}}} dx \cr
& = 2\mathop {\lim }\limits_{b \to + \infty } \int_3^b {\frac{1}{{{x^2} - 1}}} dx \cr
& \cr
& {\text{integrate by tables using the formula }}\int {\frac{1}{{{x^2} - {a^2}}}dx = \frac{1}{{2a}}\ln \left| {\frac{{x - a}}{{x + a}}} \right|} + C \cr
& 2\mathop {\lim }\limits_{b \to + \infty } \int_3^b {\frac{1}{{{x^2} - 1}}} dx = 2\mathop {\lim }\limits_{b \to + \infty } \left[ {\frac{1}{{2\left( 1 \right)}}\ln \left| {\frac{{x - 1}}{{x + 1}}} \right|} \right]_3^b \cr
& = \mathop {\lim }\limits_{b \to + \infty } \left[ {\ln \left| {\frac{{x - 1}}{{x + 1}}} \right|} \right]_3^b \cr
& = \mathop {\lim }\limits_{b \to + \infty } \left[ {\ln \left| {\frac{{b - 1}}{{b + 1}}} \right| - \ln \left| {\frac{{3 - 1}}{{3 + 1}}} \right|} \right] \cr
& \cr
& {\text{simplify}} \cr
& = \mathop {\lim }\limits_{b \to + \infty } \left[ {\ln \left| {\frac{{\frac{b}{b} - \frac{1}{b}}}{{\frac{b}{b} + \frac{1}{b}}}} \right| - \ln \left| {\frac{1}{2}} \right|} \right] \cr
& = \mathop {\lim }\limits_{b \to + \infty } \left[ {\ln \left| {\frac{{1 - \frac{1}{b}}}{{1 + \frac{1}{b}}}} \right| - \ln \left| {\frac{1}{2}} \right|} \right] \cr
& \cr
& {\text{calculate the limit when }}b \to + \infty \cr
& = \ln \left| {\frac{{1 - \frac{1}{\infty }}}{{1 + \frac{1}{\infty }}}} \right| - \ln \left( {\frac{1}{2}} \right) \cr
& = \ln \left| 1 \right| - \ln \left( {\frac{1}{2}} \right) \cr
& = - \ln \left( {\frac{1}{2}} \right) \cr
& = \ln 2 \cr
& {\text{then}}{\text{,}} \cr
& {\text{The integral converges to ln2}} \cr} $$
