∫ e^x≤ft(sec x+log_e(sec x+tan x))dx is equal to: (where C is an arbi…

The correct answer is Option (4) → \( e^{x}\log_{e}(\sec x+\tan x)+\mathrm{C} \)

The integral has the standard form

    \( \displaystyle \int e^{x}\left(f(x)+f'(x)\right)dx=e^{x}f(x)+C \)

Let \( f(x)=\log_{e}(\sec x+\tan x) \). Then

    \( \displaystyle f'(x)=\frac{\sec x\tan x+\sec^{2}x}{\sec x+\tan x} \)

    \( \displaystyle =\frac{\sec x(\tan x+\sec x)}{\sec x+\tan x} \)

    \( \displaystyle =\sec x \)

So the integrand is exactly \( e^{x}\left(f'(x)+f(x)\right) \), and

    \( \displaystyle \int e^{x}\left(\sec x+\log_{e}(\sec x+\tan x)\right)dx=e^{x}\log_{e}(\sec x+\tan x)+C \)