$\int\left[\frac{\log x-1}{1+(\log x)^{2}}\right]^{2} d x=$
$\int\left[\frac{\log x-1}{1+(\log x)^{2}}\right]^{2} d x=$
$\frac{x}{(1+\log x)}+c$
$\frac{x}{1+(\log x)^{2}}+c$
$\frac{x^{2}}{1+(\log x)^{2}}+c$
$\frac{1}{1+(\log x)^{2}}+c$
Solution
$\begin{aligned}
& \int\left\{\frac{(\log x-1)}{1+(\log x)^2}\right\}^2 d x=\int \frac{(\log x)^2+1-2 \log x}{\left[(\log x)^2+1\right]^2} d x \\
& =\int \frac{(\log x)^2+1-2 x\left(\log x \cdot \frac{1}{x}\right)}{\left[(\log x)^2+1\right]^2} d x \\
& =\int \frac{d}{d x}\left[\frac{x}{(\log x)^2+1}\right] d x=\frac{x}{(\log x)^2+1}+C \\
& \Rightarrow \quad x=e^t \Rightarrow \quad d x=e^t d t \\
& \therefore \quad I=\int\left\{\frac{t-1}{1+t^2}\right\}^2 e^t d t=\int \frac{\left(1+t^2\right)-2 t}{\left(1+t^2\right)^2} \cdot e^t d t \\
& \quad=\int \frac{e^t}{1+t^2} \cdot d t-\int \frac{2 t e^t}{\left(1+t^2\right)^2} \cdot d t
\end{aligned}$
Using by parts,
$\begin{aligned}
& I=\frac{1}{1+t^2} \cdot e^t-\int \frac{-1}{\left(1+t^2\right)^2} \cdot 2 t \cdot e^t d t-\int \frac{2 t e^t}{\left(1+t^2\right)^2} d t \\
&=\frac{e^t}{1+t^2}+\int \frac{2 t e^t}{\left(1+t^2\right)^2} d t-\int \frac{2 t e^t}{\left(1+t^2\right)^2} d t \\
& \therefore \quad I=\frac{e^t}{1+t^2}+C=\frac{x}{(\log x)^2+1}+C
\end{aligned}$