The integral ∫ e^x(tan x+logsec x) dx is equal to :

The correct answer is Option (3) → \(e^x\log\sec x+C\)

The integral has the form \(\int e^x\left(f(x)+f'(x)\right)dx = e^x f(x) + C\).

Let \(f(x) = \log\sec x\). Then

    \( \displaystyle f'(x) = \frac{1}{\sec x}\cdot\sec x\tan x = \tan x \)

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

    \( \displaystyle \int e^x(\tan x+\log\sec x)\,dx = e^x\log\sec x + C \)