Answer
$$2{x^{1/2}} - 3{x^{1/3}} + 6\root 6 \of x - 6\ln \left( {\root 6 \of x + 1} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{dx}}{{\sqrt x + \root 3 \of x }}} \cr
& {\text{letting }}x = {u^6},{\text{ }}dx = 6{u^5}du \cr
& = \int {\frac{{dx}}{{\sqrt x + \root 3 \of x }}} = \int {\frac{{6{u^5}du}}{{\sqrt {{u^6}} + \root 3 \of {{u^6}} }}} \cr
& = \int {\frac{{6{u^5}du}}{{{u^3} + {u^2}}}} \cr
& = \int {\frac{{6{u^3}du}}{{u + 1}}} \cr
& {\text{by the long division}} \cr
& = \int {\left( {6{u^2} - 6u + 6 - \frac{6}{{u + 1}}} \right)} du \cr
& {\text{Integrating}} \cr
& = 2{u^3} - 3{u^2} + 6u - 6\ln \left| {u + 1} \right| + C \cr
& x = {u^6} \to u = \root 6 \of x \cr
& = 2{\left( {\root 6 \of x } \right)^3} - 3{\left( {\root 6 \of x } \right)^2} + 6\root 6 \of x - 6\ln \left| {\root 6 \of x + 1} \right| + C \cr
& = 2{x^{1/2}} - 3{x^{1/3}} + 6\root 6 \of x - 6\ln \left( {\root 6 \of x + 1} \right) + C \cr} $$
