The given differential equation is $x \frac{dy}{dx} + 2y = x^2 \log x$. To solve, we first rewrite it in the standard linear form $\frac{dy}{dx} + P(x) y = Q(x)$. Dividing both sides by $x$ ($x \neq 0$), we get:

$$\frac{dy}{dx} + \frac{2}{x} y = x \log x$$

Here, $P = \frac{2}{x}$ and $Q = x \log x$.

The integrating factor (I.F.) is:

$$e^{\int P \, dx} = e^{\int \frac{2}{x} \, dx} = e^{2 \log x} = x^2$$

Multiplying both sides of the linear equation by the I.F., we obtain:

$$x^2 \frac{dy}{dx} + 2x \, y = x^3 \log x$$

The left-hand side is the derivative of $y x^2$ with respect to $x$, i.e.,

$$\frac{d}{dx} (y x^2) = x^3 \log x$$

Integrating both sides with respect to $x$, we get:

$$y x^2 = \int x^3 \log x \, dx + C$$

To evaluate the integral, we use integration by parts. Let $u = \log x$ and $dv = x^3 \, dx$. Then $du = \frac{1}{x} dx$ and $v = \frac{x^4}{4}$. Thus,

$$\int x^3 \log x \, dx = \frac{x^4}{4} \log x - \int \frac{x^4}{4} \cdot \frac{1}{x} \, dx = \frac{x^4}{4} \log x - \frac{1}{4} \int x^3 \, dx = \frac{x^4}{4} \log x - \frac{x^4}{16}$$

Substituting back, we have:

$$y x^2 = \frac{x^4}{4} \log x - \frac{x^4}{16} + C$$

Finally, dividing by $x^2$ yields the general solution:

$$y = \frac{x^2}{4} \log x - \frac{x^2}{16} + \frac{C}{x^2}$$