1. **Problem:** Integrate with respect to $x$ the function $$\frac{23-1}{x^2 + 4x - 12}.$$ 2. **Simplify the numerator:** $$23 - 1 = 22.$$ So the integral becomes $$\int \frac{22}{x^2 + 4x - 12} \, dx.$$ 3. **Factor the denominator:** $$x^2 + 4x - 12 = (x + 6)(x - 2).$$ 4. **Rewrite the integral:** $$\int \frac{22}{(x + 6)(x - 2)} \, dx.$$ 5. **Use partial fractions:** Assume $$\frac{22}{(x + 6)(x - 2)} = \frac{A}{x + 6} + \frac{B}{x - 2}.$$ Multiply both sides by $(x + 6)(x - 2)$: $$22 = A(x - 2) + B(x + 6).$$ 6. **Find $A$ and $B$ by substituting values:** - For $x = 2$: $$22 = A(2 - 2) + B(2 + 6) = 8B \Rightarrow B = \frac{22}{8} = \frac{11}{4}.$$ - For $x = -6$: $$22 = A(-6 - 2) + B(-6 + 6) = -8A \Rightarrow A = -\frac{22}{8} = -\frac{11}{4}.$$ 7. **Rewrite integral with partial fractions:** $$\int \left( \frac{-\frac{11}{4}}{x + 6} + \frac{\frac{11}{4}}{x - 2} \right) dx = \frac{11}{4} \int \left( \frac{1}{x - 2} - \frac{1}{x + 6} \right) dx.$$ 8. **Integrate term-by-term:** $$\frac{11}{4} \left( \ln|x - 2| - \ln|x + 6| \right) + C = \frac{11}{4} \ln \left| \frac{x - 2}{x + 6} \right| + C.$$ **Final answer:** $$\boxed{\int \frac{22}{x^2 + 4x - 12} dx = \frac{11}{4} \ln \left| \frac{x - 2}{x + 6} \right| + C}.$$