Which of the following transformations reduce the differential equation $\frac{d z}{d x}+\frac{z}{x} \log z=\frac{z}{x^2}(\log z)^2$ into the form $\frac{d u}{d x}+P(x) u=Q(x)$

$u=(\log z)^{-1}$

We have,

$\frac{1}{z(\log z)^2} \frac{d z}{d x}+\frac{1}{x} \times \frac{1}{\log z}=\frac{1}{x^2}$

Putting $\frac{1}{\log z}=u$ and $-\frac{1}{z(\log z)^2} \frac{d z}{d x}=\frac{d u}{d x}$, we get

$-\frac{d u}{d x}+\frac{u}{x}=\frac{1}{x^2}$

$\Rightarrow \frac{d u}{d x}+\left(-\frac{1}{x}\right) u=-\frac{1}{x^2}$ or, $\frac{d u}{d x}+P(x) u=Q(x)$, where

$P(x)=-\frac{1}{x}$ and $Q(x)=-\frac{1}{x^2}$