## Thomas' Calculus 13th Edition

$$\frac{1}{{90}}$$
\eqalign{ & \int_0^{1/2} {{x^3}{{\left( {1 + 9{x^4}} \right)}^{ - 3/2}}} dx \cr & {\text{use the substitution method }} \cr & u = 1 + 9{x^4},\,\,\,\,du = 36{x^3}dx,\,\,\,\,\,dx = \frac{{du}}{{36{x^3}}} \cr & {\text{Then}}{\text{,}} \cr & \int {{x^3}{{\left( {1 + 9{x^4}} \right)}^{ - 3/2}}} dx = \int {{x^3}{u^{ - 3/2}}} \left( {\frac{{du}}{{36{x^3}}}} \right) \cr & = \frac{1}{{36}}\int {{u^{ - 3/2}}} du \cr & {\text{integrate}} \cr & = \frac{1}{{36}}\left( {\frac{{{u^{ - 1/2}}}}{{ - 1/2}}} \right) + C \cr & = - \frac{1}{{18}}\left( {\frac{1}{{\sqrt u }}} \right) + C \cr & {\text{write in terms of }}r;{\text{ replace }}u = 1 + 9{x^4} \cr & = - \frac{1}{{18}}\left( {\frac{1}{{\sqrt {1 + 9{x^4}} }}} \right) + C \cr & {\text{Then}}{\text{,}} \cr & \int_0^{1/2} {{x^3}{{\left( {1 + 9{x^4}} \right)}^{ - 3/2}}} dx = - \frac{1}{{18}}\left( {\frac{1}{{\sqrt {1 + 9{x^4}} }}} \right)_0^{1/2} \cr & {\text{Evaluating, we get:}} \cr & = - \frac{1}{{18}}\left( {\frac{1}{{\sqrt {1 + 9{{\left( {1/2} \right)}^4}} }}} \right) + \frac{1}{{18}}\left( {\frac{1}{{\sqrt {1 + 9{{\left( 0 \right)}^4}} }}} \right) \cr & = - \frac{1}{{18}}\left( {\frac{1}{{5/4}}} \right) + \frac{1}{{18}}\left( 1 \right) \cr & = \frac{1}{{90}} \cr}