Question:

Step 1: Identify the condition for the function to be defined. The function $f(x) = \ln(x^2 + ax + 1)$ involves a natural logarithm. For the logarithm to be defined, its argument must be strictly positive. Therefore, we must have $x^2 + ax + 1 > 0$ for all $x \in \mathbb{R}$. Step 2: Analyze the nature of the quadratic expression. The expression $x^2 + ax + 1$ is a quadratic polynomial. The coefficient of $x^2$ is $1$, which is positive. This means the parabola representing the quadratic opens upwards. For this quadratic to be strictly positive for all real values of $x$, it must not intersect or touch the x-axis. This implies that the quadratic equation $x^2 + ax + 1 = 0$ must have no real roots. Step 3: Apply the condition on the discriminant for no real roots. For a quadratic equation $Ax^2 + Bx + C = 0$ to have no real roots, its discriminant $\Delta = B^2 - 4AC$ must be strictly negative. In our quadratic expression $x^2 + ax + 1$, we have $A=1$, $B=a$, and $C=1$. Therefore, the condition for no real roots is: $$ \Delta = a^2 - 4(1)(1) < 0 $$ $$ a^2 - 4 < 0 $$ Step 4: Solve the inequality for 'a'. We need to find the values of $a$ that satisfy the inequality $a^2 - 4 < 0$. $$ a^2 < 4 $$ Taking the square root of both sides, we get: $$ \sqrt{a^2} < \sqrt{4} $$ $$ |a| < 2 $$ This inequality can be rewritten as: $$ -2 < a < 2 $$ Step 5: Determine the number of integers in the range of 'a'. The integers that lie strictly between $-2$ and $2$ are the integers greater than $-2$ and less than $2$. These integers are $-1, 0, 1$. The set of integer values for $a$ is $\{-1, 0, 1\}$. Counting these values, there are $3$ integers. Step 6: State the final answer. The number of integers in the range of 'a' is $3$. The final answer is $\boxed{3}$.