1. **Stating the problem:** We want to evaluate the integral $$\int \frac{\cos(2x)+2\sin^2(x)}{\cos^2(x)}\,dx.$$\n\n2. **Simplify the integrand:** Recall the identity $$\cos(2x) = \cos^2(x) - \sin^2(x),$$ and also $$\sin^2(x) = 1 - \cos^2(x).$$ Substitute these into the numerator:\n$$\cos(2x) + 2\sin^2(x) = (\cos^2(x) - \sin^2(x)) + 2(1 - \cos^2(x)) = \cos^2(x) - \sin^2(x) + 2 - 2\cos^2(x) = -\cos^2(x) - \sin^2(x) + 2.$$\nBut since $$\sin^2(x) + \cos^2(x) = 1,$$ it follows that:\n$$-\cos^2(x) - \sin^2(x) + 2 = -1 + 2 = 1.$$\n\n3. **Rewrite the integral:** The integrand simplifies to:\n$$\frac{\cos(2x) + 2 \sin^2(x)}{\cos^2(x)} = \frac{1}{\cos^2(x)} = \sec^2(x).$$\n\n4. **Integrate:** The integral becomes:\n$$\int \sec^2(x) \, dx = \tan(x) + C,$$ where $C$ is the constant of integration.\n\n**Final answer:** $$\int \frac{\cos(2x)+2\sin^2(x)}{\cos^2(x)} \, dx = \tan(x) + C.$$