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1)
$f(x)=\frac{1}{\sqrt[3]{(x+2)^2}+\sqrt[3]{x^2-4}+\sqrt[3]{(x-2)^2} }=\frac{\sqrt[3]{x+2}-\sqrt[3]{x-2}}{(x+2)-(x-2)}=\frac{1}{4}\sqrt[3]{x+2}-\frac{1}{4}\sqrt[3]{x-2}$
Vậy $\int f(x)dx=\int \left (\frac{1}{4}\sqrt[3]{x+2}-\frac{1}{4}\sqrt[3]{x-2} \right )dx=\frac{3}{16}\sqrt[3]{(x+2)^4}-\frac{3}{16}\sqrt[3]{(x-2)^4}+C$
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