The set of values of x for which the inequality $|x-1|+|x+1|<4$ always holds true, is

$(-2, 2)$

On the LHS of the given inequation there are two terms $|x-1|$ and $|x + 1|$. On equating $x - 1$ and $x + 1$ to zero, we get: $x = -1$ and 1 as critical points. These points divide the real line into three regions viz. $(-∞, -1), [-1, 1)$ and $[1, ∞)$. So, we consider the following cases:

CASE I When $-∞ < x < -1$

In this case, we have

$|x-1|=-(x-1)$ and $|x+1| = -(x + 1)$

$∴|x-1|+|x+1|<4$

$⇒-(x-1)-(x + 1) < 4⇒-2x <4⇒ x>-2$

But, $-∞<x<-1$

$∴x ∈ (-2, -1)$

CASE II When $-1 ≤ x <1$

In this case, we have

$|x-1|=-(x-1)$ and $|x+1|=x+1$

$∴|x-1|+|x+1|<4$

$⇒-(x-1)+x+1<4$

$⇒2 <4$, which is true for all $x ∈ [-1, 1)$.

$∴x ∈[-1,1)$

CASE III When $1 ≤x<∞$

In this case, we have

$|x-1|= x-1$ and $|x+1| = x+1$

$∴|x-1|+|x+1|<4$

$⇒x-1+x+1<4⇒ x <2$

$⇒ x ∈ [1, 2)$  $[∵1≤x<∞]$

Hence, $x ∈ (-2, -1) ∪[-1, 1)∪[1, 2)$ or, $x ∈ (-2, 2)$.