Question

The integral I = $\int e^x (\frac{x-1}{3x^2}) dx$ is equal to

• A. $\frac{1}{3}(\frac{x^2}{2} - x) + C$, where C is constant of integration
• B. $(\frac{x^2}{2} - x)e^x + C$, where C is constant of integration
• C. $\frac{1}{3x^2}e^x + C$, where C is constant of integration
• D. $\frac{1}{3x}e^x + C$, where C is constant of integration

Hint:

Whenever you see an integral involving a product of $e^x$ and another function, always try to check if the other function can be expressed as a sum of a function and its derivative, i.e., $f(x) + f'(x)$. This pattern is very common in competitive exams.

Answer 1

The Correct Option is D

Step 1: Concept Identification: The integral is in the form $\int e^x (f(x) + f'(x)) dx = e^x f(x) + C$. The objective is to transform the integrand to match this property. Step 2: Core Strategy: Identify a function $f(x)$ and its derivative $f'(x)$ within the expression. Step 3: Detailed Analysis: Rewrite the expression within the parentheses by separating the terms: \[ \frac{x-1}{3x^2} = \frac{x}{3x^2} - \frac{1}{3x^2} = \frac{1}{3x} - \frac{1}{3x^2} \] The integral is thus: \[ I = \int e^x \left(\frac{1}{3x} - \frac{1}{3x^2}\right) dx \] We now verify if this conforms to the form $\int e^x (f(x) + f'(x)) dx$. Let \(f(x) = \frac{1}{3x} = \frac{1}{3}x^{-1}\). The derivative of $f(x)$ is: \[ f'(x) = \frac{d}{dx}\left(\frac{1}{3}x^{-1}\right) = \frac{1}{3}(-1)x^{-2} = -\frac{1}{3x^2} \] This derivative precisely matches the second term in the parentheses. Therefore, the integral is in the structure $\int e^x (f(x) + f'(x)) dx$ with $f(x) = \frac{1}{3x}$. The solution to this integral is $e^x f(x) + C$. \[ I = e^x \left(\frac{1}{3x}\right) + C = \frac{e^x}{3x} + C \] Step 4: Conclusion: The integral evaluates to $\frac{1}{3x}e^x + C$.