Evaluate $\displaystyle \int \dfrac{\log x - 1}{\left(\log x\right)^2} \; dx$
$\begin{aligned}
\text{Let } I & = \int \dfrac{\log x - 1}{\left(\log x\right)^2} \; dx \\\\
& = \int \left[\dfrac{1}{\log x} - \dfrac{1}{\left(\log x\right)^2}\right] \; dx \;\;\; \cdots \; (1)
\end{aligned}$
Let $\log x = t$ $\;\;\; \cdots (2a)$
$\implies$ $x = e^t$ $\;\;\; \cdots \; (2b)$
Differentiating equation $(2a)$ gives
$\dfrac{1}{x} \; dx = dt$ $\implies$ $dx = x \; dt$ $\implies$ $dx = e^t \; dt$ $\;\;\;$ [by equation $(2a)$] $\;\;\; \cdots \; (2c)$
$\therefore$ $\;$ In view of equations $(2a)$ and $(2c)$, equation $(1)$ becomes
$I = \displaystyle \int \left[e^t \left(\dfrac{1}{t} - \dfrac{1}{t^2}\right)\right] \; dt$ $\;\;\; \cdots \; (3)$
If $f\left(t\right) = \dfrac{1}{t}$, then $f'\left(t\right) = \dfrac{-1}{t^2}$
$\therefore$ $\;$ Equation $(3)$ can be rewritten as
$I = \displaystyle \int e^t \left[\dfrac{1}{t} + \dfrac{d}{dt} \left(\dfrac{1}{t}\right)\right] \; dt$
i.e. $I = \dfrac{e^t}{t} + c = \dfrac{e^{\log x}}{\log x} + c$ $\;\;\;$ [by equation $(2a)$]
i.e. $I = \dfrac{x}{\log x} + c$
$\left[\text{Note: }\displaystyle \int e^x \left[f\left(x\right) + f'\left(x\right)\right] \; dx = e^x \; f\left(x\right) + c\right]$