1. **State the problem:** Evaluate the integral $$\int \frac{1 + 41x}{1 + x^2} \, dx.$$\n\n2. **Recall the formula and rules:** We can split the integral into two simpler integrals:\n$$\int \frac{1}{1 + x^2} \, dx + 41 \int \frac{x}{1 + x^2} \, dx.$$\nThe integral $$\int \frac{1}{1 + x^2} \, dx = \arctan(x) + C$$ and for $$\int \frac{x}{1 + x^2} \, dx,$$ use substitution.\n\n3. **Evaluate the second integral:** Let $$u = 1 + x^2,$$ then $$du = 2x \, dx,$$ so $$x \, dx = \frac{du}{2}.$$\n\n4. Substitute into the integral:\n$$41 \int \frac{x}{1 + x^2} \, dx = 41 \int \frac{1}{u} \cdot \frac{du}{2} = \frac{41}{2} \int \frac{1}{u} \, du = \frac{41}{2} \ln|u| + C = \frac{41}{2} \ln(1 + x^2) + C.$$\n\n5. **Combine results:**\n$$\int \frac{1 + 41x}{1 + x^2} \, dx = \arctan(x) + \frac{41}{2} \ln(1 + x^2) + C.$$\n\n**Final answer:** $$\boxed{\arctan(x) + \frac{41}{2} \ln(1 + x^2) + C}.$$