Evaluate $$I=\int_{1}^{2}\frac{x^5\:dx}{\sqrt[3]{3x+x^5}}$$
My try:
we have $$5I=\int_{1}^{2}\frac{5x^5\:dx}{\sqrt[3]{3x+x^5}}=\int_{1}^{2}\frac{x(5x^4+3-3)\:dx}{\sqrt[3]{3x+x^5}}$$
Using Integration by Parts we get:
$$5I=\frac{3x}{2}\sqrt[3]{(3x+x^5)^2}\:\:\vert_{1}^{2}-\frac{3}{2}\int_{1}^{2}\sqrt[3]{(3x+x^5)^2}\:dx-\int_{1}^{2}\frac{3x}{\sqrt[3]{3x+x^5}}$$
$$5I=3\sqrt[3]{(38)^2}-\frac{3}{2}\sqrt[3]{16}-\frac{3}{2}\int_{1}^{2}\sqrt[3]{(3x+x^5)^2}\:dx-\int_{1}^{2}\frac{3x}{\sqrt[3]{3x+x^5}}$$
Any way from here?