1. **Problem Statement:** Given that $\int f(x)\,dx = g(x) + c_1$ and $\int g(x)\,dx = \ln\left(\frac{1}{x}\right) + c_2$, find the value of $f(2) + g(2)$.\n\n2. **Understanding the problem:** The integral of $f(x)$ is $g(x) + c_1$, so differentiating both sides gives $f(x) = g'(x)$. Similarly, the integral of $g(x)$ is $\ln\left(\frac{1}{x}\right) + c_2$, so differentiating both sides gives $g(x) = \frac{d}{dx} \left( \ln\left(\frac{1}{x}\right) \right)$.\n\n3. **Calculate $g(x)$:**\n$$g(x) = \frac{d}{dx} \left( \ln\left(\frac{1}{x}\right) \right) = \frac{d}{dx} (-\ln x) = -\frac{1}{x}.$$\n\n4. **Calculate $f(x)$:** Since $f(x) = g'(x)$, differentiate $g(x)$:\n$$f(x) = \frac{d}{dx} \left(-\frac{1}{x}\right) = -\frac{d}{dx} \left(x^{-1}\right) = -(-1)x^{-2} = \frac{1}{x^2}.$$\n\n5. **Evaluate $f(2)$ and $g(2)$:**\n$$f(2) = \frac{1}{2^2} = \frac{1}{4}, \quad g(2) = -\frac{1}{2}.$$\n\n6. **Sum the values:**\n$$f(2) + g(2) = \frac{1}{4} - \frac{1}{2} = -\frac{1}{4}.$$\n\n**Final answer:** $f(2) + g(2) = -\frac{1}{4}$, which corresponds to option (a).