$$\int \tan ^{-1}(\sec x+\tan x) d x=$$

If $$\int \frac{1+x^2}{1+x^4} d x=\frac{1}{\sqrt{2}} \tan ^{-1}\left[\frac{f(x)}{\sqrt{2}}\right]+c$$, then $$f(x)=$$

$$\int \frac{x+\sin x}{1+\cos x} d x=$$

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 $$k=$$