$\int\left\{\frac{\log x-1}{1+(\log x)^2}\right\}^2 d x$ is equal to

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

We have,

$I =\int\left\{\frac{\log x-1}{1+(\log x)^2}\right\}^2 d x$

$\Rightarrow I =\int e^t \frac{(t-1)^2}{\left(t^2+1\right)^2} d t$, where $t=\log x$

$\Rightarrow I=\int e^t \frac{t^2+1-2 t}{\left(t^2+1\right)^2} d t$

$\Rightarrow I=\int e^t\left\{\frac{1}{t^2+1}+\frac{-2 t}{\left(t^2+1\right)^2}\right\} d t$

$\Rightarrow I=\frac{e^t}{t^2+1}+C=\frac{x}{(\log x)^2+1}+C$