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Q.
$\int(\sec x+\tan x)^{2} d x=$
Integrals
Solution:
$\int(\sec x+\tan x)^{2} d x =\int\left(\sec ^{2} x+\tan ^{2} x+2 \sec x \tan x\right) d x$
$=\int\left(2 \sec ^{2} x-1+2 \sec x \tan x\right) d x$
$=2(\sec x+\tan x)-x+ c$