[TEX] \[I = \int {\frac{{dx}}{{{{\cos }^2}(1 + \ln x)}}} = \frac{x}{{{{\cos }^2}(1 + \ln x)}} - \int {\tan (1 + \ln x)dx = } \frac{x}{{{{\cos }^2}(1 + \ln x)}} - x\tan (1 + \ln x) + \int {\frac{x}{{{{\cos }^2}(1 + \ln x)x}}dx} = \frac{x}{{{{\cos }^2}(1 + \ln x)}} - x\tan (1 + \ln x) + I = > I = \frac{1}{2}(\frac{x}{{{{\cos }^2}(1 + \ln x)}} - x\tan (1 + \ln x))\] [/TEX]