Answer
$$\frac{1}{7}$$
Work Step by Step
$$\eqalign{
& \int_1^2 {\frac{4}{{9{x^2} + 6x + 1}}} dx \cr
& = \int_1^2 {\frac{4}{{{{\left( {3x} \right)}^2} + 2\left( {3x} \right) + {{\left( 1 \right)}^2}}}} dx \cr
& {\text{factor the perfect square trinomial}} \cr
& = \int_1^2 {\frac{4}{{{{\left( {3x + 1} \right)}^2}}}} dx \cr
& {\text{set }}u = 3x + 1{\text{ then }}du = 3dx \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{1}{3}du = dx \cr
& {\text{switch the limits of integration}} \cr
& u = 3x + 1,\,\,\,\,x = 1 \to u = 4 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = 2 \to u = 7 \cr
& {\text{use the change of variable}} \cr
& \int_1^2 {\frac{4}{{{{\left( {3x + 1} \right)}^2}}}} dx = \int_4^7 {\frac{4}{{{u^2}}}} \left( {\frac{1}{3}du} \right) \cr
& = \frac{4}{3}\int_4^7 {{u^{ - 2}}} du \cr
& {\text{integrate}} \cr
& = \frac{4}{3}\left[ {\frac{{{u^{ - 1}}}}{{ - 1}}} \right]_4^7 \cr
& = \frac{4}{3}\left( { - \frac{1}{u}} \right)_4^7 \cr
& {\text{evaluate the limits}} \cr
& = - \frac{4}{3}\left( {\frac{1}{7} - \frac{1}{4}} \right) \cr
& = - \frac{4}{3}\left( { - \frac{3}{{28}}} \right) \cr
& = \frac{1}{7} \cr} $$
