Question

\( \int e^x \left(\frac{1}{x} - \frac{1}{x^2}\right) dx = ............. + c \)

Hint:

When you see an integral involving \( e^x \) multiplied by a sum or difference of functions, always check if it fits the form \( \int e^x [f(x) + f'(x)] dx \). Recognizing this pattern saves a lot of time compared to using integration by parts.

Answer 1

Step 1: Identify the integral's form. It matches the special property \( \int e^x [f(x) + f'(x)] dx = e^x f(x) + c \).
Step 2: Let \( f(x) = \frac{1}{x} \). Find its derivative, \( f'(x) \). \[ f'(x) = \frac{d}{dx}(x^{-1}) = -1 \cdot x^{-2} = -\frac{1}{x^2} \] Step 3: Confirm that the integrand matches the form \( e^x[f(x) + f'(x)] \). \[ e^x \left(\frac{1}{x} - \frac{1}{x^2}\right) = e^x [f(x) + f'(x)] \] The form is correct. Therefore, the integral is \( e^x f(x) + c \).
Step 4: Substitute back \( f(x) = \frac{1}{x} \) to get the final answer. \[ \int e^x \left(\frac{1}{x} - \frac{1}{x^2}\right) dx = \frac{e^x}{x} + c \]