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: Understanding the Concept:
This integral is in a special form that matches the property $\int e^x (f(x) + f'(x)) dx = e^x f(x) + C$. We need to manipulate the integrand to fit this form.
Step 2: Key Formula or Approach:
Identify a function f(x) and its derivative f'(x) within the parentheses.
Step 3: Detailed Explanation:
First, let's rewrite the expression inside the parentheses by splitting the fraction:
\[ \frac{x-1}{3x^2} = \frac{x}{3x^2} - \frac{1}{3x^2} = \frac{1}{3x} - \frac{1}{3x^2} \] So the integral becomes:
\[ I = \int e^x \left(\frac{1}{3x} - \frac{1}{3x^2}\right) dx \] Now let's check if this is in the form $\int e^x (f(x) + f'(x)) dx$.
Let's try setting \(f(x) = \frac{1}{3x} = \frac{1}{3}x^{-1}\).
Now find the derivative of f(x):
\[ f'(x) = \frac{d}{dx}\left(\frac{1}{3}x^{-1}\right) = \frac{1}{3}(-1)x^{-2} = -\frac{1}{3x^2} \] This matches the second term in the parentheses perfectly.
So, our integral is indeed in the form $\int e^x (f(x) + f'(x)) dx$ with $f(x) = \frac{1}{3x}$.
The result of this integration is $e^x f(x) + C$.
\[ I = e^x \left(\frac{1}{3x}\right) + C = \frac{e^x}{3x} + C \] Step 4: Final Answer:
The value of the integral is $\frac{1{3x}e^x + C$}.