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

Question

( A ) $\frac{x}{x^2+1} +c$
( B ) $\frac{\log x}{(\log x)^2 +1} + c$
( C ) $\frac{\log x}{(\log x)^2 + 1} + c$
( D ) $\frac{x}{(\log x)^2 + 1} + c$