Since the given logarithm does exist, then the base and the argument of logarithm respect the conditions of existence.

We'll impose the constraints of existence of logarithms:

1) x+1 > 0

2) x + 1 different from 1.

3) x^2-5x+6 > 0

We'll solve the first constraint:

1) x+1 > 0

We'll subtract 1 both sides:

x > -1

2) x + 1 different from 1

We'll subtract 1 both sides:

x different from -1+1 = 0

3) x^2-5x+6 > 0

The expression is positive for values of x located outside the roots of the expression x^2-5x+6 = 0.

We'll calculate the roots of the quadratic:

x1 = [5 + sqrt(25-24)]/2

x1 = (5+1)/2

x1 = 3

x2 = (5-1)/2

x2 = 2

The expression is positive if x is in the interval (-infinite ; 2)U(3 ; +infinite)

From the interval (-infinite ; 2), we'll reject the value 0.

The intervals (-infinite ; 2)U(3 ; +infinite) - {0} will be intersected by the interval (-1 ; +infinity).

**The interval of admissible values for the logarithm to be defined is (-1;2)U(3 ; +infinite) - {0}.**

To determine x if log(x+1) (x^2-5x+6) exists.

For log(x+1) (x^2-5x+6) to exist, the expression (x+1) (x^2-5x+6) > 0.

For this first we factorise x^2-5x+6.

x^2+5x+6 = (x+3)(x+2).

So (x+1)(x^2+5x+6) = (x+1)(x+3)(x+2) = (x+3)(x+2)(x+1) which is positive, if x > -1,

If -2< x< -1), then (x+1)(x+3)(x+2) < 0.

If** -3<x<-2**, then (x+1)(x+3)(x+2) > 0, as only last two factors are -ve.

If x < 3, then (x+1)(x+3)(x+2) < 0 as all 3 factors are -ve.

So (x+1) (x^2-5x+6) = (x+1)(x+3)(x+2) > o when x is in (-3 , -2) or x is in (-1 , +infinity).

Therefore log(x+1) (x^2-5x+6) exists, when x is in (-3 , -2) or x is in (-1 , +infinity).