Carry the logarithm domain through a hence inequality

Divide the inequality by the positive constant 2 to get $2/(x+1)\le|x|$. Sketch y=2/(x+1), with vertical asymptote x=−1 and horizontal asymptote y=0, and add y=|x|. The crossing input is x=1. Read where the hyperbola is on or below the V: x<−1 or x≥1. The pole −1 is excluded; the crossing 1 is included.

Divide $4/(\ln x+1)\le2|\ln x|$ by 2. It has the same form with the old input replaced by ln x, including inside the modulus. Thus ln x<−1 or ln x≥1.

The logarithm requires x>0, and its denominator excludes ln x=−1. Since ln is increasing, the intervals become $0<x<e^{-1}$ or $x\ge e$. The lower interval must begin at 0, not at−∞; the source’s common-mistake note warns that this bound is often missed. The threshold e⁻¹ is excluded and e is included.