## Thomas' Calculus 13th Edition

$$\frac{4}{3}\left( {{3^{3/2}} - {2^{3/2}}} \right)$$
\eqalign{ & \int_1^4 {\frac{{{{\left( {1 + \sqrt u } \right)}^{1/2}}}}{{\sqrt u }}} du \cr & {\text{use the substitution method }} \cr & t = 1 + \sqrt u,\,\,\,\,dt = \frac{1}{{2\sqrt u }}du,\,\,\,\,\,du = 2\sqrt u dt \cr & {\text{Then}}{\text{,}} \cr & \int {\frac{{{{\left( {1 + \sqrt u } \right)}^{1/2}}}}{{\sqrt u }}} du = \int {\frac{{{t^{1/2}}}}{{\sqrt u }}} \left( {2\sqrt u dt} \right) \cr & = 2\int {{t^{1/2}}} dt \cr & {\text{integrate}} \cr & = 2\left( {\frac{{{t^{3/2}}}}{{3/2}}} \right) + C \cr & = \frac{4}{3}{t^{3/2}} + C \cr & {\text{write in terms of }}u;{\text{ replace }}t = 1 + \sqrt u \cr & = \frac{4}{3}{\left( {1 + \sqrt u } \right)^{3/2}} + C \cr & {\text{Then}}{\text{,}} \cr & \int_1^4 {\frac{{{{\left( {1 + \sqrt u } \right)}^{1/2}}}}{{\sqrt u }}} du = \left( {\frac{4}{3}{{\left( {1 + \sqrt u } \right)}^{3/2}}} \right)_1^4 \cr & {\text{Evaluating, we get:}} \cr & = \frac{4}{3}{\left( {1 + \sqrt 4 } \right)^{3/2}} - \frac{4}{3}{\left( {1 + \sqrt 1 } \right)^{3/2}} \cr & = \frac{4}{3}{\left( 3 \right)^{3/2}} - \frac{4}{3}{\left( 2 \right)^{3/2}} \cr & = \frac{4}{3}\left( {{3^{3/2}} - {2^{3/2}}} \right) \cr}