Answer
$$\frac{\pi }{2}$$
Work Step by Step
$$\eqalign{
& \int_{ - 1}^1 {\frac{{dx}}{{1 + {x^2}}}} \cr
& {\text{integrate using the formula }}\int {\frac{{du}}{{{a^2} + {u^2}}} = \frac{1}{a}{{\tan }^{ - 1}}\frac{u}{a} + C,{\text{ }}} \left( {{\text{see page 468}}} \right) \cr
& {\text{let }}a = 1 \cr
& \int_{ - 1}^1 {\frac{{dx}}{{1 + {x^2}}}} = \left. {\left( {{{\tan }^{ - 1}}x} \right)} \right|_{ - 1}^1 \cr
& {\text{part 1 of fundamental theorem of calculus}} \cr
& = {\tan ^{ - 1}}\left( 1 \right) - {\tan ^{ - 1}}\left( { - 1} \right) \cr
& {\text{simplify}} \cr
& = \frac{\pi }{4} - \left( { - \frac{\pi }{4}} \right) \cr
& = \frac{\pi }{4} + \frac{\pi }{4} \cr
& = \frac{\pi }{2} \cr} $$
