Chapter 5 - Section 5.5 - The Substitution Rule - 5.5 Exercises - Page 426: 80

$\frac{{28}}{3} + \frac{4}{3}\ln \left( {\frac{2}{9}} \right)$

$$\eqalign{ & \int_1^{16} {\frac{{{x^{1/2}}}}{{1 + {x^{3/4}}}}} dx \cr & {\text{Let }}u = \root 4 \of x \to du = \frac{1}{4}{x^{ - 3/4}}dx,{\text{ }}dx = 4{x^{3/4}}du,{\text{ }}dx = 4{u^3}du \cr & {\text{Finding the new limits of integration}} \cr & x = 16 \to u = \root 4 \of {16} = 2 \cr & x = 1 \to u = \root 4 \of 1 = 1 \cr & {\text{Applying the substitution}}{\text{, we obtain}} \cr & \int_1^{16} {\frac{{{x^{1/2}}}}{{1 + {x^{3/4}}}}} dx = \int_1^{16} {\frac{{{{\left( {{x^{1/4}}} \right)}^2}}}{{1 + {{\left( {{x^{1/4}}} \right)}^3}}}} dx \cr & \int_1^{16} {\frac{{{{\left( {{x^{1/4}}} \right)}^2}}}{{1 + {{\left( {{x^{1/4}}} \right)}^3}}}} dx = \int_1^2 {\frac{{{u^2}}}{{1 + {u^3}}}\left( {4{u^3}} \right)du} \cr & = \int_1^2 {\frac{{4{u^5}}}{{1 + {u^3}}}du} \cr & {\text{By the long division }}\frac{{4{u^5}}}{{1 + {u^3}}} = 4{u^2} - \frac{{4{u^2}}}{{1 + {u^3}}} \cr & = \int_1^2 {\left( {4{u^2} - \frac{{4{u^2}}}{{1 + {u^3}}}} \right)du} \cr & = \int_1^2 {4{u^2}du} - \int_1^2 {\frac{{4{u^2}}}{{1 + {u^3}}}du} \cr & = 4\int_1^2 {{u^2}du} - \frac{4}{3}\int_1^2 {\frac{{3{u^2}}}{{1 + {u^3}}}du} \cr & {\text{Integrating}} \cr & = \left[ {\frac{4}{3}{u^3} - \frac{4}{3}\ln \left| {1 + {u^3}} \right|} \right]_1^2 \cr & = \left[ {\frac{4}{3}{{\left( 2 \right)}^3} - \frac{4}{3}\ln \left| {1 + {{\left( 2 \right)}^3}} \right|} \right] - \left[ {\frac{4}{3}{{\left( 1 \right)}^3} - \frac{4}{3}\ln \left| {1 + {{\left( 1 \right)}^3}} \right|} \right] \cr & {\text{Simplifying}} \cr & = \left( {\frac{{32}}{3} - \frac{4}{3}\ln \left| 9 \right|} \right) - \left( {\frac{4}{3} - \frac{4}{3}\ln \left| 2 \right|} \right) \cr & = \frac{{32}}{3} - \frac{4}{3}\ln \left| 9 \right| - \frac{4}{3} + \frac{4}{3}\ln \left| 2 \right| \cr & = \frac{{28}}{3} + \frac{4}{3}\ln \left( {\frac{2}{9}} \right) \cr} $$