The integrating factor of the differential equation $x\dfrac{dy}{dx} + 2y = x^2 \log x$ is

The given differential equation is $x\dfrac{dy}{dx} + 2y = x^2 \log x$.

Dividing the entire equation by $x$, we get:

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

This is a linear differential equation of the form $\dfrac{dy}{dx} + Py = Q$, where $P = \dfrac{2}{x}$ and $Q = x \log x$.

The integrating factor (IF) is given by $e^{\int P dx}$.

$\text{IF} = e^{\int \dfrac{2}{x} dx}$

$\text{IF} = e^{2 \log x}$

$\text{IF} = e^{\log x^2}$

$\text{IF} = x^2$

Answer: $x^2$