Question

The value of $\int e^x \left(\frac{x - 1}{x^2}\right) dx =$

• A. $-\frac{e^x}{x^2} + c$
• B. $-\frac{e^x}{x} + c$
• C. $\frac{e^x}{x^2} + c$
• D. $\frac{e^x}{x} + c$

Hint:

Whenever you see an integral containing $e^x$ multiplied by an algebraic expression, always prioritize splitting terms or adding/subtracting values to match the structural form $\int e^x(f(x) + f'(x))dx$. It eliminates complex integration by parts and yields the answer instantly.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We need to solve an indefinite integral involving an exponential function multiplied by an algebraic fraction.

Step 2: Key Formula or Approach:
We look for an arrangement to match the standard special exponential integral rule: $$\int e^x \Big( f(x) + f'(x) \Big) dx = e^x f(x) + c$$ We will split the fractional term to separate it into a function and its respective derivative.

Step 3: Detailed Explanation:
Let us rewrite and simplify the given integrand expression: $$I = \int e^x \left(\frac{x - 1}{x^2}\right) dx$$ Divide each term in the numerator by the denominator $x^2$: $$I = \int e^x \left(\frac{x}{x^2} - \frac{1}{x^2}\right) dx$$ $$I = \int e^x \left(\frac{1}{x} - \frac{1}{x^2}\right) dx$$ Let $f(x) = \frac{1}{x} = x^{-1}$. Differentiating $f(x)$ with respect to $x$ using the power rule: $$f'(x) = -1 \cdot x^{-2} = -\frac{1}{x^2}$$ Now substitute these assignments back into our integral template format: $$I = \int e^x \Big( f(x) + f'(x) \Big) dx$$ Applying the theorem directly: $$I = e^x f(x) + c = e^x \left(\frac{1}{x}\right) + c = \frac{e^x}{x} + c$$

Step 4: Final Answer:
The calculated integral matches option (D).