\[ I = \int \left( \frac{\log x - 1}{1 + (\log x)^2} \right)^2 dx \] Let \(\log x = t\), so that \( x = e^t \) and \( dx = e^t dt \). \[ I = \int \left( \frac{t - 1}{1 + t^2} \right)^2 e^t dt \] \[ = \int e^t \left( \frac{t^2 + 1 - 2t}{(1 + t^2)^2} \right) dt \] \[ = \int e^t \left( \frac{1}{t^2 + 1} - \frac{2t}{(1 + t^2)^2} \right) dt \quad \text{...(1)} \] Using the integral formula: \[ \int e^x (f(x) + f'(x)) dx = e^x f(x) + C \quad \text{...(2)} \] From equation (1), we set: \[ f(x) = \frac{1}{t^2 + 1}, \quad \text{so that} \quad f'(x) = \frac{-2t}{(t^2 + 1)^2} \] Thus, applying equation (2): \[ I = e^t \left( \frac{1}{t^2 + 1} \right) + C = e^{\log x} \left( \frac{1}{(\log x)^2 + 1} \right) + C \] \[ I = x \left( \frac{1}{(\log x)^2 + 1} \right) + C \]