To identify which rational expression has a domain restriction of \( x \neq -1 \), we need to find which one has a factor in the denominator that equals zero when \( x = -1 \).
Let's analyze each option:
-
\( \frac{x^2 + 6x + 5}{6x + 6} \):
- The denominator \( 6x + 6 = 6(x + 1) \) equals zero when \( x = -1 \). This expression has a restriction of \( x \neq -1 \).
-
\( \frac{x + 1}{x^3} \):
- The denominator \( x^3 \) equals zero at \( x = 0 \). There is no restriction from \( x = -1 \).
-
\( \frac{6x^3 + 60}{x} \):
- The denominator \( x \) equals zero at \( x = 0 \). There is no restriction from \( x = -1 \).
-
\( \frac{\sqrt{x}}{2x + 2} \):
- The denominator \( 2x + 2 = 2(x + 1) \) also equals zero when \( x = -1 \), but the square root in the numerator imposes a restriction of \( x \geq 0 \) for real values to be defined.
Only the first option, \( \frac{x^2 + 6x + 5}{6x + 6} \) contains a restriction of \( x \neq -1 \) based on the denominator. Thus, the answer is:
\( \frac{x^2 + 6x + 5}{6x + 6} \).