Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.7 Exercises - Page 380: 49

$$\eqalign{ & \left( {\text{a}} \right)\frac{2}{3}{\left( {x - 1} \right)^{3/2}} + C \cr & \left( {\text{b}} \right)\frac{2}{{15}}{\left( {x - 1} \right)^{3/2}}\left( {3x + 2} \right) + C \cr & \left( {\text{c}} \right)\frac{2}{3}\sqrt {x - 1} \left( {x + 2} \right) + C \cr} $$

$$\eqalign{ & \left( {\text{a}} \right)\int {\sqrt {x - 1} } dx \cr & {\text{Integrate by substitution, let }}u = x - 1,{\text{ }}du = dx \cr & \int {\sqrt {x - 1} } dx = \int {\sqrt u } du \cr & = \frac{2}{3}{u^{3/2}} + C \cr & {\text{write in terms of }}x \cr & = \frac{2}{3}{\left( {x - 1} \right)^{3/2}} + C \cr & \cr & \left( {\text{b}} \right)\int {x\sqrt {x - 1} } dx \cr & {\text{Integrate by substitution, let }}u = x - 1,{\text{ }}x = u + 1,{\text{ }}du = dx \cr & \int {x\sqrt {x - 1} } dx = \int {\left( {u + 1} \right)\sqrt u } du \cr & = \int {\left( {{u^{3/2}} + {u^{1/2}}} \right)} du \cr & = \frac{2}{5}{u^{5/2}} + \frac{2}{3}{u^{3/2}} + C \cr & = \frac{2}{{15}}{u^{3/2}}\left( {3u + 5} \right) + C \cr & {\text{write in terms of }}x \cr & = \frac{2}{{15}}{\left( {x - 1} \right)^{3/2}}\left( {3\left( {x - 1} \right) + 5} \right) + C \cr & = \frac{2}{{15}}{\left( {x - 1} \right)^{3/2}}\left( {3x + 2} \right) + C \cr & \cr & \left( {\text{c}} \right)\int {\frac{x}{{\sqrt {x - 1} }}} dx \cr & {\text{Integrate by substitution, let }}u = x - 1,{\text{ }}x = u + 1,{\text{ }}du = dx \cr & \int {\frac{x}{{\sqrt {x - 1} }}} dx = \int {\frac{{u + 1}}{{\sqrt u }}} du \cr & = \int {\left( {{u^{1/2}} + {u^{ - 1/2}}} \right)} du \cr & = \frac{2}{3}{u^{3/2}} + 2{u^{1/2}} + C \cr & = \frac{2}{3}{u^{1/2}}\left( {u + 3} \right) + C \cr & {\text{write in terms of }}x \cr & = \frac{2}{3}\sqrt {x - 1} \left( {x - 1 + 3} \right) + C \cr & = \frac{2}{3}\sqrt {x - 1} \left( {x + 2} \right) + C \cr} $$