Answer
$$\int(\sec^2x)e^{\tan x}dx=e^{\tan x}+C$$
Work Step by Step
$$A=\int(\sec^2x)e^{\tan x}dx$$
We set $a=\tan x$, which means $$da=\sec^2xdx$$
Therefore, $$A=\int e^ada$$ $$A=e^a+C$$ $$A=e^{\tan x}+C$$
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