Answer
$$3\ln \left( {\frac{{2 + \sqrt 5 }}{{1 + \sqrt 2 }}} \right)$$
Work Step by Step
$$\eqalign{
& \int_{1/8}^1 {\frac{{dx}}{{x\sqrt {1 + {x^{2/3}}} }}} \cr
& {\text{or we can write as}}: \cr
& = \int_{1/8}^1 {\frac{{dx}}{{x\sqrt {1 + {{\left( {{x^{1/3}}} \right)}^2}} }}} \cr
& {\text{set }}u = {x^{1/3}}{\text{ }}\,\,\,\,\,{\text{then }}\,\,\,\,du = \frac{1}{3}{x^{ - 2/3}}dx,\,\,\,\,\,\,du = \frac{1}{3}{x^{ - 1/3}}{x^{ - 1}}dx \cr
& or\,\,\,\,\,\,\,\,\,\,\,\,\frac{{3du}}{{{x^{1/3}}}} = \frac{1}{x}dx \to \frac{{3du}}{u} = \frac{1}{x}dx \cr
& {\text{switch the limits of integration}} \cr
& u = {x^{1/3}},\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = 1/8 \to u = 1/2 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = 1 \to u = 1 \cr
& {\text{use the change of variable}} \cr
& \int_{1/8}^1 {\frac{{dx}}{{x\sqrt {1 + {{\left( {{x^{1/3}}} \right)}^2}} }}} = \int_{1/2}^1 {\frac{{3du}}{{u\sqrt {1 + {u^2}} }}} \cr
& = 3\int_{1/2}^1 {\frac{{du}}{{u\sqrt {1 + {u^2}} }}} \cr
& {\text{Using the Integral formulas in the Theorem 6}}{\text{.12}}:{\text{ }} \cr
& \int {\frac{{dx}}{{x\sqrt {{a^2} + {x^2}} }} = - \frac{1}{a}{{\operatorname{csch} }^{ - 1}}\frac{{\left| x \right|}}{a} + C,{\text{ for }}x \ne 0} {\text{ Then}}{\text{,}} \cr
& = 3\int_{1/2}^1 {\frac{{du}}{{u\sqrt {1 + {u^2}} }}} = - 3\left( {{{\operatorname{csch} }^{ - 1}}\left| x \right|} \right)_{1/2}^1 \cr
& = - 3\left( {{{\operatorname{csch} }^{ - 1}}\left| 1 \right| - {{\operatorname{csch} }^{ - 1}}\left| {1/2} \right|} \right) \cr
& = 3{\operatorname{csch} ^{ - 1}}\left( {1/2} \right) - 3{\operatorname{csch} ^{ - 1}}\left( 1 \right) \cr
& {\text{Using the theorem 6}}{\text{.10 to express the answer in terms of logarithms}}:{\text{ }} \cr
& {\operatorname{csch} ^{ - 1}}x = {\cosh ^{ - 1}}\left( {\frac{1}{x}} \right){\text{ for }}x \ne 0.{\text{ Then}}{\text{,}} \cr
& = 3{\sinh ^{ - 1}}\left( 2 \right) - 3{\sinh ^{ - 1}}\left( 1 \right) \cr
& {\text{And from the theorem 6}}{\text{.10 we have that }}{\sinh ^{ - 1}}x = \ln \left( {x + \sqrt {{x^2} + 1} } \right){\text{ then}}{\text{,}} \cr
& = 3\ln \left( {2 + \sqrt {{2^2} + 1} } \right) - 3\ln \left( {1 + \sqrt {{1^2} + 1} } \right) \cr
& = 3\ln \left( {2 + \sqrt 5 } \right) - 3\ln \left( {1 + \sqrt 2 } \right) \cr
& = 3\ln \left( {\frac{{2 + \sqrt 5 }}{{1 + \sqrt 2 }}} \right) \cr} $$
