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

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

$=\int [\frac{1}{(1+(\log x{)}^2)}-\frac{2\log x}{{(1+(\log x{)}^2)}^2}]dx$

$=\int [\frac{e^t}{1+t^2}-\frac{2t{e}^t}{{(1+t^2)}^2}]dt \text{put }\log x=t\Rightarrow dx=e^{t}dt$

$\int e^t[\frac{1}{1+t^2}-\frac{2t}{{(1+t^2)}^2}]dt$

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