Answer
$$\frac{n}{{b\left( {1 + n} \right)}}{u^{\frac{{1 + n}}{n}}} + C$$
Work Step by Step
$$\eqalign{
& \int {\root n \of {a + bx} } dx \cr
& {\text{substitute }}u = a + bx,{\text{ }}du = bdx \cr
& \int {\root n \of {a + bx} } dx = \int {\root n \of u } \left( {\frac{1}{b}du} \right) \cr
& = \frac{1}{b}\int {{u^{1/n}}du} \cr
& {\text{find antiderivative }} \cr
& = \frac{1}{b}\left( {\frac{{{u^{1/n + 1}}}}{{1/n + 1}}} \right) + C \cr
& {\text{simplify}} \cr
& = \frac{1}{b}\left( {\frac{{{u^{\frac{{1 + n}}{n}}}}}{{\frac{{1 + n}}{n}}}} \right) + C \cr
& = \frac{n}{{b\left( {1 + n} \right)}}{u^{\frac{{1 + n}}{n}}} + C \cr} $$
