Calculus: Early Transcendentals (2nd Edition)

$$\frac{\pi }{9}$$
\eqalign{ & \int_{1/3}^{1/\sqrt 3 } {\frac{4}{{9{x^2} + 1}}dx} \cr & {\text{substitute }}u = 3x,{\text{ }}du = 3dx \cr & {\text{express the limits in terms of }}u \cr & x = 1/3{\text{ implies }}u = 3\left( {1/3} \right) = 1 \cr & x = 1/\sqrt 3 {\text{ implies }}u = 3\left( {1/\sqrt 3 } \right) = 3/\sqrt 3 \cr & {\text{the entire integration is carried out as follows}} \cr & \int_{1/3}^{1/\sqrt 3 } {\frac{4}{{9{x^2} + 1}}dx} = 4\int_1^{3/\sqrt 3 } {\frac{{1/3}}{{{u^2} + 1}}} du \cr & = \frac{4}{3}\int_1^{3/\sqrt 3 } {\frac{{du}}{{{u^2} + 1}}} \cr & {\text{find the antiderivative}} \cr & = \frac{4}{3}\left. {\left( {{{\tan }^{ - 1}}u} \right)} \right|_1^{3/\sqrt 3 } \cr & {\text{use the fundamental theorem}} \cr & = \frac{4}{3}\left( {{{\tan }^{ - 1}}\left( {\frac{3}{{\sqrt 3 }}} \right) - {{\tan }^{ - 1}}\left( 1 \right)} \right) \cr & {\text{Simplify}} \cr & = \frac{4}{3}\left( {\frac{\pi }{3} - \frac{\pi }{4}} \right) \cr & = \frac{\pi }{9} \cr}