$$\int\left[-\frac{\log x-1}{1+(\log x)^2}\right]^2 d x=$$

$$\int \frac{d x}{\cos 2 x-\cos ^2 x}=$$

$$\int \frac{1+2 e^{-x}}{1-2 e^{-x}} d x=$$

If $$\int \frac{\sin \theta}{\sin 3 \theta} d \theta=\frac{1}{2 k} \log \left|\frac{k+\tan \theta}{k-\tan \theta}\right|+c$$, then $$k=$$