Answer
$$2{x^{1/2}} - 3{x^{1/3}} + 6{x^{1/3}} - 6\ln \left( {{x^{1/6}} + 1} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{dx}}{{\sqrt x + \root 3 \of x }}} \cr
& = \int {\frac{{dx}}{{{x^{1/2}} + {x^{1/3}}}}} \cr
& {\text{Taking the hint on the book making a substitution of the form }}u = {x^{1/n}} \cr
& {\text{Then}}{\text{, the common factor for }}\frac{1}{2}{\text{ and }}\frac{1}{3}{\text{ is }}\frac{1}{6} \cr
& {\text{Take }}\,\,\,u = {x^{1/6}} \to \,\,\,\,\,\,\,\,{u^6} = x,\,\,\,\,\,\,\,\,6{u^5}du = dx\,\,\,\, \cr
& \cr
& {\text{write the integrand in terms of }}u \cr
& \int {\frac{{dx}}{{{x^{1/2}} + {x^{1/3}}}}} = \int {\frac{{6{u^5}du}}{{{{\left( {{u^6}} \right)}^{1/2}} + {{\left( {{u^6}} \right)}^{1/3}}}}} \cr
& = \int {\frac{{6{u^5}du}}{{{u^3} + {u^2}}}} \cr
& = \int {\frac{{6{u^5}du}}{{{u^2}\left( {u + 1} \right)}}} \cr
& = \int {\frac{{6{u^3}du}}{{u + 1}}} \cr
& \cr
& {\text{perform the long division}} \cr
& \frac{{6{u^3}}}{{u + 1}} = 6{u^2} - 6u + 6 - \frac{6}{{u + 1}} \cr
& \int {\frac{{6{u^3}du}}{{u + 1}}} = \int {\left( {6{u^2} - 6u + 6 - \frac{6}{{u + 1}}} \right)} du \cr
& \cr
& {\text{integrating}} \cr
& = 6\left( {\frac{{{u^3}}}{3}} \right) - 6\left( {\frac{{{u^2}}}{2}} \right) + 6u - 6\ln \left| {u + 1} \right| + C \cr
& = 2{u^3} - 3{u^2} + 6u - 6\ln \left| {u + 1} \right| + C \cr
& {\text{write the integrand in terms of }}x;{\text{ replace }}u = {x^{1/6}} \cr
& = 2{\left( {{x^{1/6}}} \right)^3} - 3{\left( {{x^{1/6}}} \right)^2} + 6\left( {{x^{1/6}}} \right) - 6\ln \left| {{x^{1/6}} + 1} \right| + C \cr
& = 2{x^{1/2}} - 3{x^{1/3}} + 6{x^{1/3}} - 6\ln \left( {{x^{1/6}} + 1} \right) + C \cr} $$
