$\int\left(\frac{\log x-1}{1+(\log x)^2}\right)^2 d x=$
$\int\left(\frac{\log x-1}{1+(\log x)^2}\right)^2 d x=$
$\frac{\log x}{1+(\log x)^2}+c$
$\frac{x}{x^2+1}+c$
$\frac{x}{1+(\log x)^2}+c$
$\frac{-x}{1+(\log x)^2}+c$
Solution
We have,
$
\begin{aligned}
& I=\int\left\{\frac{\log x-1}{1+(\log x)^2}\right\}^2 d x \\
&=\int e^t \frac{(t-1)^2}{\left(t^2+1\right)^2} d t, \quad \text { where } t=\log x \\
&=\int e^t \frac{t^2+1-2 t}{\left(t^2+1\right)^2} d t \\
&=\int e^t\left\{\frac{1}{t^2+1}+\frac{-2 t}{\left(t^2+1\right)^2}\right\} d t \\
&=\frac{e^t}{t^2+1}+C=\frac{x}{(\log x)^2+1}+C .
\end{aligned}
$