$\int \sec x . \log (\sec x + \tan x) d x =$

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${(\log (\sec x + \tan x))}^{2} + c$

$\frac{1}{2} {(\log (\sec x + \tan x))}^{2} + c$

$\frac{1}{3} {(\log (\sec x + \tan x))}^{2} + c$

$\frac{1}{2} {(\log (\sec x - \tan x))}^{2} + c$

answer is C.

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Detailed Solution

$\begin{array}{l} I = \int \sec x . \log (\sec x + \tan x) d x \\ \log (\sec x + \tan x) = t \\ \Rightarrow \sec x . d x = d t \\ I = \int t d t = \frac{t^{2}}{2} + c \\ I = \frac{{[\log (\sec x + \tan x)]}^{2}}{2} + c \end{array}$

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