1. **Problem Statement:** Calculate the integral $$\int \frac{1}{(x-1)(x+2)} \, dx$$. 2. **Formula and Method:** We use partial fraction decomposition for integrals of rational functions where the denominator factors into linear terms. 3. **Partial Fraction Setup:** Express $$\frac{1}{(x-1)(x+2)}$$ as $$\frac{A}{x-1} + \frac{B}{x+2}$$. 4. **Find A and B:** Multiply both sides by $$(x-1)(x+2)$$: $$1 = A(x+2) + B(x-1)$$ 5. **Solve for A and B:** Set $x=1$: $$1 = A(1+2) + B(0) \Rightarrow 1 = 3A \Rightarrow A = \frac{1}{3}$$ Set $x=-2$: $$1 = A(0) + B(-2-1) \Rightarrow 1 = -3B \Rightarrow B = -\frac{1}{3}$$ 6. **Rewrite the integral:** $$\int \frac{1}{(x-1)(x+2)} \, dx = \int \left( \frac{1/3}{x-1} - \frac{1/3}{x+2} \right) dx = \frac{1}{3} \int \frac{1}{x-1} dx - \frac{1}{3} \int \frac{1}{x+2} dx$$ 7. **Integrate:** $$\frac{1}{3} \ln|x-1| - \frac{1}{3} \ln|x+2| + C$$ 8. **Final answer:** $$\boxed{\frac{1}{3} \ln\left| \frac{x-1}{x+2} \right| + C}$$ This completes the solution with detailed steps.