Answer
$$40{y^{1/4}} - \frac{{3{y^{4/3}}}}{4} + 8{y^{1/2}} + C$$
Work Step by Step
$$\eqalign{
& \int {\left[ {\frac{{10}}{{{y^{3/4}}}} - \root 3 \of y + \frac{4}{{\sqrt y }}} \right]dy} \cr
& {\text{Rewrite the integrand}} \cr
& = \int {\left( {\frac{{10}}{{{y^{3/4}}}} - {y^{1/3}} + \frac{4}{{{y^{1/2}}}}} \right)dy} \cr
& = \int {\left( {10{y^{ - 3/4}} - {y^{1/3}} + 4{y^{ - 1/2}}} \right)dy} \cr
& {\text{Sum rule}} \cr
& = \int {10{y^{ - 3/4}}dy} - \int {{y^{1/3}}dy} + \int {4{y^{ - 1/2}}dy} \cr
& {\text{Constant rule}} \cr
& = 10\int {{y^{ - 3/4}}dy} - \int {{y^{1/3}}dy} + 4\int {{y^{ - 1/2}}dy} \cr
& {\text{Integrate by using the power rule}} \cr
& = 10\left( {\frac{{{y^{ - 3/4 + 1}}}}{{ - 3/4 + 1}}} \right) - \frac{{{y^{1/3 + 1}}}}{{1/3 + 1}} + 4\left( {\frac{{{y^{ - 1/2 + 1}}}}{{ - 1/2 + 1}}} \right) + C \cr
& = 10\left( {\frac{{{y^{1/4}}}}{{1/4}}} \right) - \frac{{{y^{4/3}}}}{{4/3}} + 4\left( {\frac{{{y^{1/2}}}}{{1/2}}} \right) + C \cr
& = 40{y^{1/4}} - \frac{{3{y^{4/3}}}}{4} + 8{y^{1/2}} + C \cr} $$
