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$\displaystyle F(x) = \int_2^x \sqrt{t-1} dt$ then $\displaystyle F(x^2) = \int_2^{x^2} \sqrt{t-1}dt$. Thus, $\displaystyle [F(x^2)]' = 2xF'(x^2) = 2x\sqrt{x^2-1}$.
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