1. **State the problem:** We need to find the indefinite integral $$\int \left(e^{3x} + \frac{2}{1-x}\right) dx.$$\n\n2. **Recall the integral rules:**\n- The integral of a sum is the sum of the integrals: $$\int (f(x) + g(x)) dx = \int f(x) dx + \int g(x) dx.$$\n- For exponential functions, $$\int e^{ax} dx = \frac{1}{a} e^{ax} + C.$$\n- For rational functions of the form $$\int \frac{1}{1-x} dx,$$ use substitution or recognize it as a logarithmic integral.\n\n3. **Split the integral:**\n$$\int e^{3x} dx + \int \frac{2}{1-x} dx.$$\n\n4. **Integrate the first term:**\nUsing the formula for exponential integrals,\n$$\int e^{3x} dx = \frac{1}{3} e^{3x} + C_1.$$\n\n5. **Integrate the second term:**\nRewrite the integral as $$2 \int \frac{1}{1-x} dx.$$\nUse substitution: let $$u = 1 - x,$$ then $$du = -dx,$$ so $$dx = -du.$$\nThe integral becomes\n$$2 \int \frac{1}{u} (-du) = -2 \int \frac{1}{u} du = -2 \ln|u| + C_2 = -2 \ln|1-x| + C_2.$$\n\n6. **Combine the results:**\n$$\int \left(e^{3x} + \frac{2}{1-x}\right) dx = \frac{1}{3} e^{3x} - 2 \ln|1-x| + C,$$\nwhere $$C = C_1 + C_2$$ is the constant of integration.