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 an exponential function $e^{ax}$ is $$\int e^{ax} dx = \frac{1}{a} e^{ax} + C.$$\n- The integral of a rational function like $\frac{1}{1-x}$ can be handled by substitution.\n\n3. **Split the integral:**\n$$\int \left(e^{3x} + \frac{2}{1-x}\right) dx = \int e^{3x} dx + \int \frac{2}{1-x} dx.$$\n\n4. **Integrate the first term:**\nUsing the formula for exponential integrals,\n$$\int e^{3x} dx = \frac{1}{3} e^{3x} + C.$$\n\n5. **Integrate the second term:**\nLet $u = 1 - x$, then $du = -dx$, so $dx = -du$.\nRewrite the integral:\n$$\int \frac{2}{1-x} dx = \int \frac{2}{u} (-du) = -2 \int \frac{1}{u} du = -2 \ln|u| + C = -2 \ln|1-x| + C.$$\n\n6. **Combine results:**\n$$\int \left(e^{3x} + \frac{2}{1-x}\right) dx = \frac{1}{3} e^{3x} - 2 \ln|1-x| + C.$$\n\n**Final answer:** $$\boxed{\frac{1}{3} e^{3x} - 2 \ln|1-x| + C}.$$