1. **State the problem:** Evaluate the integral $$\int \frac{1}{x^2 \ln x} \, dx.$$\n\n2. **Substitution:** Let $$t = \ln x,$$ then $$dt = \frac{1}{x} dx,$$ so $$dx = x dt.$$\n\n3. **Rewrite the integral:** Since $$x = e^t,$$ we have $$x^2 = e^{2t}.$$ The integral becomes\n$$\int \frac{1}{x^2 \ln x} dx = \int \frac{1}{e^{2t} t} \cdot x dt = \int \frac{1}{e^{2t} t} \cdot e^t dt = \int \frac{e^t}{e^{2t} t} dt = \int \frac{1}{e^t t} dt = \int \frac{e^{-t}}{t} dt.$$\n\n4. **Result:** The integral $$\int \frac{e^{-t}}{t} dt$$ does not have an elementary antiderivative and is expressed in terms of the Exponential Integral function $$\text{Ei}(-t).$$\n\n5. **Final answer:**\n$$\int \frac{1}{x^2 \ln x} dx = -\text{Ei}(-\ln x) + C,$$ where $$C$$ is the constant of integration.\n\n**Note:** The Exponential Integral function $$\text{Ei}(z)$$ is a special function defined as $$\text{Ei}(z) = -\int_{-z}^\infty \frac{e^{-t}}{t} dt$$ for real $$z < 0.$$