An estimate in the introduction of the Hilbert transform in Grafakos's Classical Fourier Analysis

Question

In the introduction of the Hilbert transform on the Schwartz space, the following estimate is made in Grafakos's Classical Fourier Analysis (3rd) (Section 5.1.1):

Question: Would anyone explain how the second term in (5.1.2) is done?

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It should come from the second term of (5.1.1) according to the discussion between the two formulas. But I don't see how.

Answer 1

$$\int_{|x|\ge 1}\frac{\phi(x)}{x} dx=\int_{|x|\ge 1}\frac{x\phi(x)}{x^2} dx \le \sup_{x\in \Bbb R}|x\phi(x)|\int_{|x|\ge 1}\frac{1}{x^2} dx= 2\sup_{x\in \Bbb R}|x\phi(x)|$$ since

$$\int_{|x|\ge 1}\frac{1}{x^2} dx= 2\int_{x\ge 1}\frac{1}{x^2} dx= 2\left[\frac{-1}{x}\right]^{\infty}_{1} =2$$