Prove

Question:

$\int \sec x(\sec x+\tan x) d x$

Solution:

$\int \sec x(\sec x+\tan x) d x$

$=\int\left(\sec ^{2} x+\sec x \tan x\right) d x$

$=\int \sec ^{2} x d x+\int \sec x \tan x d x$

$=\tan x+\sec x+\mathrm{C}$

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