1. **State the problem:** We need to evaluate the definite integral $$\int_{2.1}^{2.3} \left(\cos x - \ln x + e^{-x}\right) \, dx.$$\n\n2. **Recall the integral rules:** The integral of a sum is the sum of the integrals: $$\int (f(x) + g(x) + h(x)) \, dx = \int f(x) \, dx + \int g(x) \, dx + \int h(x) \, dx.$$\nAlso, the integral of each function is: $$\int \cos x \, dx = \sin x + C,$$$$\int \ln x \, dx = x \ln x - x + C,$$$$\int e^{-x} \, dx = -e^{-x} + C.$$\n\n3. **Set up the integral:**\n$$\int_{2.1}^{2.3} \cos x \, dx - \int_{2.1}^{2.3} \ln x \, dx + \int_{2.1}^{2.3} e^{-x} \, dx.$$\n\n4. **Evaluate each integral separately:**\n- For $$\int_{2.1}^{2.3} \cos x \, dx = \sin x \Big|_{2.1}^{2.3} = \sin 2.3 - \sin 2.1.$$\n- For $$\int_{2.1}^{2.3} \ln x \, dx = \left(x \ln x - x\right) \Big|_{2.1}^{2.3} = (2.3 \ln 2.3 - 2.3) - (2.1 \ln 2.1 - 2.1).$$\n- For $$\int_{2.1}^{2.3} e^{-x} \, dx = \left(-e^{-x}\right) \Big|_{2.1}^{2.3} = -e^{-2.3} + e^{-2.1} = e^{-2.1} - e^{-2.3}.$$\n\n5. **Calculate numerical values:**\n- $$\sin 2.3 \approx 0.7457,$$$$\sin 2.1 \approx 0.8632,$$$$\sin 2.3 - \sin 2.1 \approx 0.7457 - 0.8632 = -0.1175.$$\n- $$2.3 \ln 2.3 \approx 2.3 \times 0.8329 = 1.9157,$$$$2.3 \ln 2.3 - 2.3 = 1.9157 - 2.3 = -0.3843,$$$$2.1 \ln 2.1 \approx 2.1 \times 0.7419 = 1.5579,$$$$2.1 \ln 2.1 - 2.1 = 1.5579 - 2.1 = -0.5421,$$$$\text{difference} = -0.3843 - (-0.5421) = 0.1578.$$\n- $$e^{-2.1} \approx 0.1225,$$$$e^{-2.3} \approx 0.1003,$$$$e^{-2.1} - e^{-2.3} = 0.1225 - 0.1003 = 0.0222.$$\n\n6. **Sum all parts:**\n$$-0.1175 - 0.1578 + 0.0222 = -0.2531.$$\n\n**Final answer:** $$\int_{2.1}^{2.3} \left(\cos x - \ln x + e^{-x}\right) \, dx \approx -0.253.$$