Answer
$$\int x\sqrt[3]{x+c}dx=\frac{3}{7}(x+c)^{\frac{7}{3}}-\frac{3}{4}c(x+c)^{\frac{4}{3}}+C$$
Work Step by Step
$$let\,t = \sqrt[3]{x+c},dx=3t^{2}dt$$
$$\int x\sqrt[3]{x+c}dx=3\int (t^{6}-ct^{3})dt$$
$$=\frac{3}{7}t^{7}-\frac{3}{4}ct^{4}+C$$
$$=\frac{3}{7}(x+c)^{\frac{7}{3}}-\frac{3}{4}c(x+c)^{\frac{4}{3}}+C$$