Answer
$$\frac{{6{x^{7/6}}}}{7} + \frac{{3{x^{2/3}}}}{2} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{\sqrt x + 1}}{{\root 3 \of x }}} dx \cr
& {\text{rewrite the radicals using }}\root n \of x = {x^{1/n}} \cr
& = \int {\frac{{{x^{1/2}} + 1}}{{{x^{1/3}}}}} dx \cr
& {\text{distributive property}} \cr
& = \int {\left( {\frac{{{x^{1/2}}}}{{{x^{1/3}}}} + \frac{1}{{{x^{1/3}}}}} \right)} dx \cr
& {\text{simplifying}} \cr
& = \int {\left( {{x^{1/6}} + \frac{1}{{{x^{1/3}}}}} \right)} dx \cr
& = \int {\left( {{x^{1/6}} + {x^{ - 1/3}}} \right)} dx \cr
& {\text{use }}\int {{x^n}dx} = \frac{{{x^{n + 1}}}}{{n + 1}} + C{\text{ }} \cr
& = \frac{{{x^{1/6 + 1}}}}{{1/6 + 1}} + \frac{{{x^{ - 1/3 + 1}}}}{{ - 1/3 + 1}} + C \cr
& = \frac{{{x^{7/6}}}}{{7/6}} + \frac{{{x^{2/3}}}}{{2/3}} + C \cr
& = \frac{{6{x^{7/6}}}}{7} + \frac{{3{x^{2/3}}}}{2} + C \cr} $$
