The integral I = ∫ e^x≤ft((x-1)/(3x^2)) dx is equal to

The correct answer is Option (4) → \(\displaystyle\frac{1}{3x}e^x + C\), where C is constant of integration

Splitting the fraction:

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

With \(f(x) = \frac{1}{x}\), \(f'(x) = -\frac{1}{x^2}\), so the bracket is \(f(x) + f'(x)\). By the standard result \(\int e^x[f(x) + f'(x)]\,dx = e^x f(x) + C\):

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

    \(\displaystyle = \frac{1}{3} \cdot \frac{e^x}{x} + C = \frac{1}{3x}e^x + C\)