## Calculus, 10th Edition (Anton)

$$\frac{{2\sqrt 3 \pi }}{3} - \frac{\pi }{6} - \frac{2}{3} + \frac{1}{3}\ln 2$$
\eqalign{ & \int_1^3 {\sqrt x {{\tan }^{ - 1}}\sqrt x } dx \cr & {\text{substitute }}u = {\tan ^{ - 1}}\sqrt x ,{\text{ }} \cr & du = \frac{{1/2\sqrt x }}{{1 + {{\left( {\sqrt x } \right)}^2}}}dx \cr & du = \frac{1}{{2\sqrt x \left( {1 + x} \right)}}dx \cr & dv = \sqrt x dx,{\text{ }}v = \frac{2}{3}{x^{3/2}} \cr & {\text{applying integration by parts}}{\text{, we have}} \cr & \int_1^3 {\sqrt x {{\tan }^{ - 1}}\sqrt x } dx = \left. {\left( {\frac{2}{3}{x^{3/2}}{{\tan }^{ - 1}}\sqrt x } \right)} \right|_1^3 - \int_1^3 {\left( {\frac{2}{3}{x^{3/2}}\frac{1}{{2\sqrt x \left( {1 + x} \right)}}dx} \right)} \cr & \int_1^3 {\sqrt x {{\tan }^{ - 1}}\sqrt x } dx = \left. {\left( {\frac{2}{3}{x^{3/2}}{{\tan }^{ - 1}}\sqrt x } \right)} \right|_1^3 - \frac{1}{3}\int_1^3 {\left( {\frac{x}{{1 + x}}dx} \right)} \cr & {\text{long division}} \cr & \int_1^3 {\sqrt x {{\tan }^{ - 1}}\sqrt x } dx = \left. {\left( {\frac{2}{3}{x^{3/2}}{{\tan }^{ - 1}}\sqrt x } \right)} \right|_1^3 - \frac{1}{3}\int_1^3 {\left( {1 - \frac{1}{{x + 1}}} \right)} dx \cr & {\text{integrating}} \cr & \int_1^3 {\sqrt x {{\tan }^{ - 1}}\sqrt x } dx = \left. {\left( {\frac{2}{3}{x^{3/2}}{{\tan }^{ - 1}}\sqrt x } \right)} \right|_1^3 - \frac{1}{3}\left. {\left( {x - \ln \left| {x + 1} \right|} \right)} \right|_1^3 \cr & \int_1^3 {\sqrt x {{\tan }^{ - 1}}\sqrt x } dx = \left. {\left( {\frac{2}{3}{x^{3/2}}{{\tan }^{ - 1}}\sqrt x - \frac{1}{3}x + \frac{1}{3}\ln \left| {x + 1} \right|} \right)} \right|_1^3 \cr & {\text{evaluate limits}} \cr & = \left( {\frac{2}{3}{{\left( 3 \right)}^{3/2}}{{\tan }^{ - 1}}\sqrt 3 - \frac{1}{3}\left( 3 \right) + \frac{1}{3}\ln \left| {3 + 1} \right|} \right) - \left( {\frac{2}{3}{{\left( 1 \right)}^{3/2}}{{\tan }^{ - 1}}\sqrt 1 - \frac{1}{3}\left( 1 \right) + \frac{1}{3}\ln \left| {1 + 1} \right|} \right) \cr & {\text{simplify}} \cr & = \left( {2\sqrt 3 \left( {\frac{\pi }{3}} \right) - 1 + \frac{1}{3}\ln 4} \right) - \left( {\frac{2}{3}\left( {\frac{\pi }{4}} \right) - \frac{1}{3} + \frac{1}{3}\ln \left| 2 \right|} \right) \cr & = \frac{{2\sqrt 3 \pi }}{3} - 1 + \frac{1}{3}\ln 4 - \frac{\pi }{6} + \frac{1}{3} - \frac{1}{3}\ln 2 \cr & = \frac{{2\sqrt 3 \pi }}{3} - \frac{\pi }{6} - \frac{2}{3} + \frac{1}{3}\ln 2 \cr}