Answer
$$\frac{{4{{\left( {\root 4 \of x } \right)}^3}}}{3} - 4\root 4 \of x + 4{\tan ^{ - 1}}\root 4 \of x + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{x^{1/4}}}}{{1 + {x^{1/2}}}}} dx \cr
& = \int {\frac{{{x^{1/4}}}}{{1 + {{\left( {{x^{1/4}}} \right)}^2}}}} dx \cr
& {\text{Let }}{u^4} = x,{\text{ 4}}{u^3}du = dx \cr
& {\text{Substituting}} \cr
& \int {\frac{{{x^{1/4}}}}{{1 + {{\left( {{x^{1/4}}} \right)}^2}}}} dx = \int {\frac{{{{\left( {{u^4}} \right)}^{1/4}}}}{{1 + {{\left( {{{\left( {{u^4}} \right)}^{1/4}}} \right)}^2}}}} \left( {4{u^3}} \right)du \cr
& = \int {\frac{u}{{1 + {u^2}}}} \left( {4{u^3}} \right)du \cr
& = 4\int {\frac{{4{u^4}}}{{1 + {u^2}}}} du \cr
& {\text{By long division}} \cr
& = 4\int {\left( {{u^2} - 1 + \frac{1}{{1 + {u^2}}}} \right)} du \cr
& = 4\int {{u^2}} du - 4\int {du} + 4\int {\frac{1}{{1 + {u^2}}}} du \cr
& {\text{Integrating}} \cr
& {\text{ = }}\frac{{4{u^3}}}{3} - 4u + 4{\tan ^{ - 1}}u + C \cr
& {\text{Write in terms of }}x,{\text{ substitute}}\root 4 \of x {\text{ for }}u \cr
& {\text{ = }}\frac{{4{{\left( {\root 4 \of x } \right)}^3}}}{3} - 4\root 4 \of x + 4{\tan ^{ - 1}}\root 4 \of x + C \cr} $$
