Answer
$$\frac{{4{{\left( {x + 2} \right)}^{3/4}}}}{3} - 2{\left( {x + 2} \right)^{1/2}} + 4{\left( {x + 2} \right)^{1/4}} - \ln {\left( {{{\left( {x + 2} \right)}^{1/4}} + 1} \right)^4} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{dx}}{{\root 4 \of {x + 2} + 1}}} \cr
& {\text{letting }}x + 2 = {u^4},{\text{ }}dx = 4{u^3}du \cr
& = \int {\frac{{dx}}{{x - \root 3 \of x }}} = \int {\frac{{3{u^2}du}}{{{u^3} - \root 3 \of {{u^3}} }}} \cr
& = \int {\frac{{dx}}{{\root 4 \of {x + 2} + 1}}} = \int {\frac{{4{u^3}du}}{{\root 4 \of {{u^4}} + 1}}} \cr
& = \int {\frac{{4{u^3}}}{{u + 1}}} du \cr
& {\text{by the long division}} \cr
& = \int {\left( {4{u^2} - 4u + 4 - \frac{4}{{u + 1}}} \right)} du \cr
& {\text{Integrating}} \cr
& = \frac{{4{u^3}}}{3} - 2{u^2} + 4u - 4\ln \left| {u + 1} \right| + C \cr
& x + 2 = {u^4} \to u = \root 4 \of {x + 2} \cr
& = \frac{{4{{\left( {\root 4 \of {x + 2} } \right)}^3}}}{3} - 2{\left( {\root 4 \of {x + 2} } \right)^2} + 4\root 4 \of {x + 2} - 4\ln \left| {\root 4 \of {x + 2} + 1} \right| + C \cr
& {\text{Simplify}} \cr
& = \frac{{4{{\left( {x + 2} \right)}^{3/4}}}}{3} - 2{\left( {x + 2} \right)^{1/2}} + 4{\left( {x + 2} \right)^{1/4}} - \ln {\left( {{{\left( {x + 2} \right)}^{1/4}} + 1} \right)^4} + C \cr} $$
