Answer
$$\frac{\pi }{4}$$
Work Step by Step
$$\eqalign{
& \int_{ - 2}^1 {\frac{3}{{{x^2} + 4x + 13}}} dx \cr
& {\text{completing the square}} \cr
& = \int_{ - 2}^1 {\frac{3}{{{x^2} + 4x + 4 + 9}}} dx \cr
& = \int_{ - 2}^1 {\frac{3}{{\left( {{x^2} + 4x + 4} \right) + 9}}} dx \cr
& = \int_{ - 2}^1 {\frac{3}{{{{\left( {x + 2} \right)}^2} + {3^2}}}} dx \cr
& {\text{find the antiderivative}} \cr
& = 3\left[ {\frac{1}{3}{{\tan }^{ - 1}}\left( {\frac{{x + 2}}{3}} \right)} \right]_{ - 2}^1 \cr
& = \left( {{{\tan }^{ - 1}}\left( {\frac{{x + 2}}{3}} \right)} \right)_{ - 2}^1 \cr
& {\text{evaluate limits}} \cr
& = \left( {{{\tan }^{ - 1}}\left( {\frac{{1 + 2}}{3}} \right) - {{\tan }^{ - 1}}\left( {\frac{{ - 2 + 2}}{3}} \right)} \right) \cr
& {\text{simplify}} \cr
& = \left( {{{\tan }^{ - 1}}\left( 1 \right) - {{\tan }^{ - 1}}\left( 0 \right)} \right) \cr
& = \frac{\pi }{4} \cr} $$
