Answer
$$2\sqrt x + 3\root 3 \of x + 6\root 6 \of x + 6\ln \left| {\root 6 \of x - 1} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{1}{{\sqrt x - \root 3 \of x }}} dx \cr
& {\text{Let }}{t^6} = x,{\text{ }}dx = 6{t^5}dt \cr
& {\text{Substituting}} \cr
& \int {\frac{1}{{\sqrt x - \root 3 \of x }}} dx = \int {\frac{1}{{\sqrt {{t^6}} - \root 3 \of {{t^6}} }}} \left( {6{t^5}dt} \right) \cr
& = \int {\frac{{6{t^5}}}{{{t^3} - {t^2}}}} dt \cr
& = \int {\frac{{6{t^3}}}{{t - 1}}} dt \cr
& {\text{By long division }}\frac{{6{t^3}}}{{t - 1}} = 6{t^2} + 6t + 6 + \frac{6}{{t - 1}} \cr
& = \int {\left( {6{t^2} + 6t + 6 + \frac{6}{{t - 1}}} \right)} dt \cr
& {\text{Integrating}} \cr
& = 2{t^3} + 3{t^2} + 6t + 6\ln \left| {t - 1} \right| + C \cr
& {\text{Write in terms of }}x,{\text{ substitute }}\root 6 \of x {\text{ for }}t \cr
& = 2{\left( {\root 6 \of x } \right)^3} + 3{\left( {\root 6 \of x } \right)^2} + 6\left( {\root 6 \of x } \right) + 6\ln \left| {\root 6 \of x - 1} \right| + C \cr
& = 2\sqrt x + 3\root 3 \of x + 6\root 6 \of x + 6\ln \left| {\root 6 \of x - 1} \right| + C \cr} $$
