Question
$\int_{}^{} {\sec x\log (\sec x + \tan x)\;dx = } $
Therefore $\int_{}^{} {\sec x\,\log (\sec x + \tan x)\,dx} = \int_{}^{} {t\,dt} $$ = \frac{{{t^2}}}{2} + c = \frac{{{{[\log (\sec x + \tan x)]}^2}}}{2} + c.$
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