Answer
$2\sqrt x + 6\ln \left( {\root 4 \of x + 1} \right) - 4\root 4 \of x + C$
Work Step by Step
$$\eqalign{
& \int {\frac{{dx}}{{{x^{1/2}} + {x^{1/4}}}}} \cr
& {\text{Let }}{u^4} = x,{\text{ }}4{u^3}du = dx \cr
& {\text{Substituting}} \cr
& \int {\frac{{dx}}{{{x^{1/2}} + {x^{1/4}}}}} = \int {\frac{1}{{{{\left( {{u^4}} \right)}^{1/2}} + {{\left( {{u^4}} \right)}^{1/4}}}}} \left( {4{u^3}} \right)du \cr
& = \int {\frac{{4{u^3}}}{{{u^2} + u}}} du \cr
& {\text{By the long division }}\frac{{4{u^3}}}{{{u^2} + u}} = 4u + \frac{4}{{u + 1}} - 4,{\text{ so}} \cr
& \int {\frac{{4{u^3}}}{{{u^2} + u}}} du = \int {\left( {4u + \frac{4}{{u + 1}} - 4} \right)} du \cr
& = \int {4u} du + \int {\frac{4}{{u + 1}}} du - \int {4du} \cr
& {\text{Integrate}} \cr
& = 2{u^2} + 6\ln \left| {u + 1} \right| - 4u + C \cr
& {\text{Write the result in terms of }}x,{\text{ }}{u^4} = x \to u = \root 4 \of x \cr
& = 2{\left( {\root 4 \of x } \right)^2} + 6\ln \left| {\root 4 \of x + 1} \right| - 4\root 4 \of x + C \cr
& {\text{Simplifying}} \cr
& = 2\sqrt x + 6\ln \left( {\root 4 \of x + 1} \right) - 4\root 4 \of x + C \cr} $$
