Official solution
\[\int {{{{{\left( {\log x - 1} \right)}^2}} \over {{{\left( {1 + {{\left( {\log x} \right)}^2}} \right)}^2}}}} dx\]
\[ = \int {{{1 + {{\left( {\log x} \right)}^2} - 2\log x} \over {{{\left[ {1 + {{\left( {\log x} \right)}^2}} \right]}^2}}}} \]
\[ = \int {\left[ {{1 \over {\left( {1 + {{\left( {\log x} \right)}^2}} \right)}} - {{2\log x} \over {{{\left( {1 + {{\left( {\log x} \right)}^2}} \right)}^2}}}} \right]} dx\]
\[ = \int {\left[ {{{{e^t}} \over {1 + {t^2}}} - {{2t\,{e^t}} \over {{{\left( {1 + {t^2}} \right)}^2}}}} \right]} dt\]
put \[\log x = t \Rightarrow dx = {e^t}\,dt\]
\[ = \int {{e^t}} \left[ {{1 \over {1 + {t^2}}} - {{2t} \over {{{\left( {1 + {t^2}} \right)}^2}}}} \right]dt\]
\[\left[ \, \right.\] which is of the form
\[\left. {\int {{e^x}\left( {f\left( x \right) + f'\left( x \right)dx} \right)} } \right]\]
\[ = {{{e^t}} \over {1 + {t^2}}} + c = {x \over {1 + {{\left( {\log x} \right)}^2}}} + c\]