1. **State the problem:** We need to solve the integral $$\int \frac{\sqrt{1+4x^2}}{x} \, dx.$$\n\n2. **Recall the formula and substitution:** This integral involves a square root of a quadratic expression. A useful substitution is to let $$2x = \sinh(t)$$ because $$\sqrt{1+\sinh^2(t)} = \cosh(t)$$. This simplifies the square root.\n\n3. **Perform the substitution:** Let $$2x = \sinh(t) \Rightarrow x = \frac{\sinh(t)}{2}$$ and $$dx = \frac{\cosh(t)}{2} dt.$$\n\n4. **Rewrite the integral:** Substitute into the integral:\n$$\int \frac{\sqrt{1+4x^2}}{x} dx = \int \frac{\sqrt{1+\sinh^2(t)}}{\frac{\sinh(t)}{2}} \cdot \frac{\cosh(t)}{2} dt = \int \frac{\cosh(t)}{\frac{\sinh(t)}{2}} \cdot \frac{\cosh(t)}{2} dt = \int \frac{\cosh^2(t)}{\sinh(t)} dt.$$\n\n5. **Simplify the integral:**\n$$\int \frac{\cosh^2(t)}{\sinh(t)} dt = \int \frac{1 + \sinh^2(t)}{\sinh(t)} dt = \int \left( \frac{1}{\sinh(t)} + \sinh(t) \right) dt.$$\n\n6. **Split the integral:**\n$$\int \frac{1}{\sinh(t)} dt + \int \sinh(t) dt.$$\n\n7. **Integrate each part:**\n- $$\int \sinh(t) dt = \cosh(t) + C.$$\n- $$\int \frac{1}{\sinh(t)} dt = \int \csch(t) dt = \ln \left| \tanh \frac{t}{2} \right| + C.$$\n\n8. **Combine results:**\n$$\int \frac{\cosh^2(t)}{\sinh(t)} dt = \ln \left| \tanh \frac{t}{2} \right| + \cosh(t) + C.$$\n\n9. **Back-substitute to x:** Recall $$t = \sinh^{-1}(2x)$$ and $$\cosh(t) = \sqrt{1 + \sinh^2(t)} = \sqrt{1 + 4x^2}.$$ Also,\n$$\tanh \frac{t}{2} = \frac{\sinh(t)}{1 + \cosh(t)} = \frac{2x}{1 + \sqrt{1 + 4x^2}}.$$\n\n10. **Final answer:**\n$$\int \frac{\sqrt{1+4x^2}}{x} dx = \ln \left| \frac{2x}{1 + \sqrt{1 + 4x^2}} \right| + \sqrt{1 + 4x^2} + C.$$