Question

Evaluate: ${\int }_3^9\frac{\sqrt[3]{312-x}}{\sqrt[3]{3x}+\sqrt[3]{312-x}}dx$

Answer 1

$I={\int }_3^9\frac{\sqrt[3]{312-x}}{\sqrt[3]{3x}+\sqrt[3]{312-x}}dx={\int }_3^9\frac{\sqrt[3]{312-(9+3-x)}}{\sqrt[3]{39+3-x}+\sqrt[3]{312-(9+3-x)}}dx$

$I={\int }_3^9\frac{\sqrt[3]{312-x}}{\sqrt[3]{3x}+\sqrt[3]{312-x}}dx={\int }_3^9\frac{\sqrt[3]{3x}}{\sqrt[3]{312-x}+\sqrt[3]{3x}}dx$

$2I={\int }_3^9\frac{\sqrt[3]{312-x}}{\sqrt[3]{3x}+\sqrt[3]{312-x}}dx+{\int }_3^9\frac{\sqrt[3]{3x}}{\sqrt[3]{312-x}+\sqrt[3]{3x}}dx$

$2I={\int }_3^{9}dx=6⟹I=3$

${\int }_3^9\frac{\sqrt[3]{312-x}}{\sqrt[3]{3x}+\sqrt[3]{312-x}}dx=3$