Answer
$$\frac{{5{{\left( {x + 3} \right)}^{9/5}}}}{9} - \frac{{3{{\left( {x + 3} \right)}^{4/5}}}}{4} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{x}{{{{\left( {x + 3} \right)}^{1/5}}}}} dx \cr
& {\text{Taking the hint on the book making a substitution of the form }}u = {\left( {x + a} \right)^{1/n}} \cr
& {\text{Take }}\,\,\,u = {\left( {x + 3} \right)^{1/5}} \to \,\,\,\,\,\,\,\,{u^5} = x + 3,\,\,\,\,\,\,\,5{u^4}du = dx\,\,\,\, \cr
& \cr
& {\text{write the integrand in terms of }}u \cr
& \int {\frac{x}{{{{\left( {x + 3} \right)}^{1/5}}}}} dx = \int {\frac{{{u^5} - 3}}{u}} \left( {5{u^4}du} \right) \cr
& = 5\int {\left( {{u^8} - 3{u^3}} \right)} du \cr
& \cr
& {\text{integrating}} \cr
& = \frac{{5{u^9}}}{9} - \frac{{3{u^4}}}{4} + C \cr
& \cr
& {\text{write the integrand in terms of }}x;{\text{ replace }}u = {\left( {x + 3} \right)^{1/5}} \cr
& = \frac{{5{{\left( {{{\left( {x + 3} \right)}^{1/5}}} \right)}^9}}}{9} - \frac{{3{{\left( {{{\left( {x + 3} \right)}^{1/5}}} \right)}^4}}}{4} + C \cr
& = \frac{{5{{\left( {x + 3} \right)}^{9/5}}}}{9} - \frac{{3{{\left( {x + 3} \right)}^{4/5}}}}{4} + C \cr} $$
