<s>[INST] $ \int_{0}^{1}\frac{\sqrt[3]{x}}{\sqrt[3]{1+x}+\sqrt[3]{1-x}}dx<0.5$ Prove:
       $ \int_{0}^{1}\frac{\sqrt[3]{x}}{\sqrt[3]{1+x}+\sqrt[3]{1-x}}dx<0.5$
My solution:
$( x+1>x)$
$$ \int_{0}^{1}\frac{\sqrt[3]{x}}{\sqrt[3]{1+x}+\sqrt[3]{1-x}}dx< \int_{0}^{1}\frac{\sqrt[3]{x}}{\sqrt[3]{x}+\sqrt[3]{1-x}}=0.5
$$
Is there a closed formula  for the integrals
 [/INST] Multiply numerator and denominator by $$\sqrt[3]{(1+x)}^2
-\sqrt[3]{1+x}\sqrt[3]{1-x}+\sqrt[3]{(1-x)^2}$$
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