$\int \frac{\sec^2 x}{1 + \tan x} \, dx = $
- A$\log (\cos x + \sin x) + c$
- B$\log (\sec^2 x) + c$
- C$\log (1 + \tan x) + c$
- D$-\frac{1}{(1 + \tan x)^2} + c$
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