Answer is: option1

Solution:

We complete the square in the denominator:

\[ - x^2 - 2x + 3 = -\left(x^2 + 2x - 3\right) = -\left((x + 1)^2 - 4\right) = 4 - (x + 1)^2 \]

So the integral becomes:

\[ \int \frac{12}{\sqrt{4 - (x + 1)^2}} \, dx \]

Let:

\[ u = \frac{x + 1}{2} \Rightarrow x + 1 = 2u, \quad dx = 2 \, du \]

Now substitute into the integral:

\[ \int \frac{12}{\sqrt{4 - (x + 1)^2}} \, dx = \int \frac{12}{\sqrt{4 - (2u)^2}} \cdot 2 \, du = \int \frac{24}{\sqrt{4 - 4u^2}} \, du = \int \frac{24}{2 \sqrt{1 - u^2}} \, du = \int \frac{12}{\sqrt{1 - u^2}} \, du \]

We know that:

\[ \int \frac{1}{\sqrt{1 - u^2}} \, du = \sin^{-1}(u) \]

So:

\[ \int \frac{12}{\sqrt{1 - u^2}} \, du = 12 \sin^{-1}(u) + C \]

Now substitute \( u = \frac{x + 1}{2} \) back:

\[ = 12 \sin^{-1} \left( \frac{x + 1}{2} \right) + C \]

Hence option A