$\int\left[\frac{1-\log x}{1+(\log x)^{2}}\right]^{2} d x=$
$\int\left[\frac{1-\log x}{1+(\log x)^{2}}\right]^{2} d x=$
$\frac{1}{1+(\log x)^{2}}+c$
$\frac{x}{1+(\log x)^{2}}+c$
$\frac{1}{1+\log x}+c$
$\frac{x}{1+\log x}+c$
Solution
$\begin{aligned}{\text{Let}} I &=\int\left[\frac{1-\log x}{1+(\log x)^{2}}\right]^{2} d x \\ \text { Put } \log x &=t \Rightarrow \frac{1}{x} d x=d t \Rightarrow d x=x \text { dt i.e. } d x=e^{t} d t \\ \therefore I &=\int e^{t}\left[\frac{1-t}{1+t^{2}}\right]^{2} d t=\int e^{t} \frac{(1-t)^{2}}{\left(1+t^{2}\right)^{2}} d t=\int e^{t}\left[\frac{1+t^{2}-2 t}{\left(1+t^{2}\right)^{2}}\right] d t \\ I &=\int e^{t}\left[\frac{1}{1+t^{2}}-\frac{2 t}{\left(1+t^{2}\right)^{2}}\right] d t=\frac{e^{t}}{1+t^{2}}+c \\ I &=\frac{x}{1+(\log x)^{2}}+c \end{aligned}$