\iint_{D^{'}}\frac{rdr d\varphi }{r^{2}\cdot \left ( 1+\sqrt[5]{r^{2}} \right )}=\int_{0}^{2\pi }d\varphi \cdot \int_{2}^{3}\frac{rdr}{r^{2}\cdot (1+\sqrt[5]{r^{2})}}