## Calculus: Early Transcendentals (2nd Edition)

$$\frac{1}{{10}}{\tan ^{ - 1}}\frac{{4z}}{5} + C$$
\eqalign{ & \int {\frac{2}{{16{z^2} + 25}}} dz \cr & {\text{take out the constant}} \cr & = 2\int {\frac{1}{{16{z^2} + 25}}} dz \cr & {\text{set }}u = 4z,{\text{ }}du = 4dz \cr & = 2\int {\frac{1}{{16{z^2} + 25}}} dz = 2\int {\frac{{\frac{1}{4}du}}{{{{\left( u \right)}^2} + {{\left( 5 \right)}^2}}}} \cr & = \frac{1}{2}\int {\frac{{du}}{{{u^2} + {{\left( 5 \right)}^2}}}} \cr & {\text{from the table 4}}{\text{.10 }}\int {\frac{{du}}{{{a^2} + {u^2}}}} = \frac{1}{a}{\tan ^{ - 1}}\frac{u}{a} + C \cr & = \frac{1}{2}\left( {\frac{1}{5}{{\tan }^{ - 1}}\frac{u}{5}} \right) + C \cr & where{\text{ }}u = 4z \cr & = \frac{1}{2}\left( {\frac{1}{5}{{\tan }^{ - 1}}\frac{{4z}}{5}} \right) + C \cr & = \frac{1}{{10}}{\tan ^{ - 1}}\frac{{4z}}{5} + C \cr & {\text{check by differentiation}} \cr & {\text{ = }}\frac{d}{{dz}}\left( {\frac{1}{{10}}{{\tan }^{ - 1}}\frac{{4z}}{5} + C} \right) \cr & {\text{ = }}\frac{1}{{10}}\frac{d}{{dz}}\left( {{{\tan }^{ - 1}}\frac{{4z}}{5}} \right) + \frac{d}{{dz}}\left( C \right) \cr & {\text{ = }}\frac{1}{{10}}\left( {\frac{5}{{{{\left( {4z} \right)}^2} + {{\left( 5 \right)}^2}}}} \right) + 0 \cr & {\text{ = }}\frac{2}{{16{z^2} + 25}} \cr}