1. **State the problem:** Evaluate the integral $$\int \frac{\sec\left(\frac{1}{v}\right) \tan\left(\frac{1}{v}\right)}{v^2} \, dv.$$\n\n2. **Identify substitution:** Let $$u = \frac{1}{v}.$$ Then, $$du = -\frac{1}{v^2} dv,$$ which implies $$dv = -v^2 du.$$\n\n3. **Rewrite the integral using substitution:** Substitute $$\frac{1}{v} = u$$ and $$\frac{1}{v^2} dv = -du$$ into the integral:\n$$\int \frac{\sec(u) \tan(u)}{v^2} dv = \int \sec(u) \tan(u) \left(-du\right) = -\int \sec(u) \tan(u) du.$$\n\n4. **Recall the integral formula:** The integral of $$\sec(u) \tan(u)$$ with respect to $$u$$ is $$\sec(u) + C.$$\n\n5. **Integrate:**\n$$-\int \sec(u) \tan(u) du = -\sec(u) + C.$$\n\n6. **Back-substitute:** Replace $$u$$ with $$\frac{1}{v}$$ to get the final answer:\n$$-\sec\left(\frac{1}{v}\right) + C.$$