The domain of sec^-1 (2x+1), is

The domain of $sec^{-1}x$ is [-∞, -1]∪ [1, ∞].

Therefore, $sec^{-1}(2x+1)$ is meaningful, if

$2x + 1 ≥ 1$  or,  $2x + 1 ≤ -1$

$⇒ 2x ≥ 0 $ or, $2x ≤ -2$

$⇒x≥0   $ or,   $x ≤-1 $

$⇒ x \, ∈ (-∞, -1] ∪ [0, ∞)$

Hence, the domain of $ sec^{-1}(2x+1) $ is $(-∞, -1] ∪ [0, ∞)$.