The value of $\int \frac{\log (x / e)}{(\log x)^2} d x$ is :

$\frac{x+1}{(\log x)^2}+c$ $\frac{x-1}{(\log x)^2}+c$ $\frac{x}{\log x}+c$ $\frac{\log x}{x}+c$

$\frac{x}{\log x}+c$

Let $I=\int \frac{\ln (x / e)}{(\ln x)^2} d x=\int \frac{\ln (x)-1}{(\ln x)^2} d x$ Put $\ln x=t \Rightarrow x=e^t \Rightarrow d x=e^t d t$ $I=\int e^t\left(\frac{t-1}{t^2}\right) d t=\int e^t\left(\frac{1}{t}-\frac{1}{t^2}\right) d t=\frac{e^{t}}{t}+c=\frac{x}{\ln x}+c$ Hence (3) is the correct answer.