∫ e^x≤ft(displaystyle1/x-2/(x^3))dx =

The correct answer is Option (1) → \(e^x\left(\displaystyle\frac{x+1}{x^2}\right)+C\)

The integral uses the result \(\int e^x\left(f(x) + f'(x)\right)dx = e^x f(x) + C\). Adding and subtracting \(\frac{1}{x^2}\) inside the bracket:

    \( \displaystyle \frac{1}{x}-\frac{2}{x^3} = \left(\frac{1}{x}+\frac{1}{x^2}\right) + \left(-\frac{1}{x^2}-\frac{2}{x^3}\right) \)

With \(f(x) = \frac{1}{x}+\frac{1}{x^2}\):

    \( \displaystyle f'(x) = -\frac{1}{x^2}-\frac{2}{x^3} \)

So the bracket is exactly \(f(x) + f'(x)\), and

    \( \displaystyle \int e^x\left(\frac{1}{x}-\frac{2}{x^3}\right)dx = e^x\left(\frac{1}{x}+\frac{1}{x^2}\right)+C \)

    \( \displaystyle = e^x\left(\frac{x+1}{x^2}\right)+C \)