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$\int (\frac{\sec x}{\log (\sec x + \tan x)}) d x =$

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2 \log (\sec x + \tan x) + c

\log (\log (\sec x + \tan x)) + c

\log \sec x + \log \tan x + c

- \log (\sec x + \tan x) + c

answer is B.

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Detailed Solution

\int \frac{\sec x}{\log (\sec x + \tan x)} d x

let

f (x) = \log (\sec x + \tan x)

then

$f^{1} (x) = \sec x$

$\int \frac{\sec x}{\log (\sec x + \tan x)} d x$ $= \log (\log \sec x + \tan x) + c$

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