## Thomas' Calculus 13th Edition

$$\frac{2}{3}{\tan ^{ - 1}}\left( {\frac{{{{\tan }^{ - 1}}\sqrt x }}{3}} \right) + C$$
\eqalign{ & \int_{}^{} {\frac{1}{{\sqrt x \left( {x + 1} \right)\left( {{{\left( {{{\tan }^{ - 1}}\sqrt x } \right)}^2} + 9} \right)}}} dx \cr & {\text{use the substitution method}}{\text{.}} \cr & u = {\tan ^{ - 1}}\sqrt x,{\text{ so that }}du = \frac{{\frac{1}{{2\sqrt x }}}}{{1 + {{\left( {\sqrt x } \right)}^2}}}dx,\,\,\,\,\,\,dx = 2\sqrt x \left( {1 + x} \right)du \cr & {\text{Write the integrand in terms of }}u \cr & \int_{}^{} {\frac{1}{{\sqrt x \left( {x + 1} \right)\left( {{{\left( {{{\tan }^{ - 1}}\sqrt x } \right)}^2} + 9} \right)}}} dx = \int_{}^{} {\frac{1}{{\sqrt x \left( {x + 1} \right)\left( {{u^2} + 9} \right)}}} \left( {2\sqrt x \left( {1 + x} \right)} \right)du \cr & {\text{cancel the common terms}} \cr & = \int_{}^{} {\frac{1}{{{u^2} + 9}}} \left( 2 \right)du \cr & = 2\int_{}^{} {\frac{1}{{{u^2} + 9}}} du \cr & {\text{integrate}} \cr & = 2\left( {\frac{1}{3}{{\tan }^{ - 1}}\left( {\frac{u}{3}} \right)} \right) + C \cr & = \frac{2}{3}{\tan ^{ - 1}}\left( {\frac{u}{3}} \right) + C \cr & {\text{Write the integrand in terms of }}u,\cr &{\text{ replace }}u = {\tan ^{ - 1}}\sqrt x \cr & = \frac{2}{3}{\tan ^{ - 1}}\left( {\frac{{{{\tan }^{ - 1}}\sqrt x }}{3}} \right) + C \cr}