(C) Let $I = \int \frac{\sec^2 x}{1 + \tan x} \, dx$. Substitute $t = 1 + \tan x$. Then,the derivative is $dt = \sec^2 x \, dx$. Substituting these in

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(C) Let $I = \int \frac{\sec^2 x}{1 + \tan x} \, dx$. Substitute $t = 1 + \tan x$. Then,the derivative is $dt = \sec^2 x \, dx$. Substituting these into the integral,we get: $I = \int \frac{1}{t} \, dt$. Integrating with respect to $t$,we obtain: $I = \log |t| + c$. Replacing $t$ with $1 + \tan x$,we get: $I = \log |1 + \tan x| + c$.