Question

If \(\int \frac{(1 + x \log x)}{xe^{-x}} dx = e^x f(x) + C\), where C is constant of integration, then f(x) is

• A. \(e^x\)
• B. \(\log x\)
• C. \(\frac{1}{x}\)
• D. \(\frac{1}{x^2}\)

Hint:

Whenever you see an integral involving \(e^x\) multiplied by a function, always try to check if the function can be expressed as the sum of another function and its derivative, i.e., \(g(x) + g'(x)\). This shortcut can save a lot of time compared to using integration by parts.

Answer 1

The Correct Option is B

Step 1: Conceptual Understanding:
This problem requires recognizing a specific integration pattern. The formula \(\int e^x (g(x) + g'(x)) dx = e^x g(x) + C\) is a standard result derived from the product rule. The objective is to transform the given integrand into this format.
Step 2: Core Formula:
The relevant formula is:\[ \int e^x (g(x) + g'(x)) dx = e^x g(x) + C \]Our task is to rewrite the integral to align with this structure.
Step 3: Detailed Derivation:
The integral to evaluate is:\[ I = \int \frac{(1 + x \log x)}{xe^{-x}} dx \]Simplify the integrand by moving \(e^{-x}\) to the numerator:\[ I = \int e^x \frac{(1 + x \log x)}{x} dx \]Separate the terms in the fraction:\[ I = \int e^x \left(\frac{1}{x} + \frac{x \log x}{x}\right) dx \]\[ I = \int e^x \left(\frac{1}{x} + \log x\right) dx \]Compare this to the form \(\int e^x (g(x) + g'(x)) dx\).
Let \(g(x) = \log x\).
Then, \(g'(x) = \frac{1}{x}\).
Substituting these into the integral yields:\[ I = \int e^x (g'(x) + g(x)) dx \]This matches the standard form precisely.
Consequently, the integration result is:\[ I = e^x g(x) + C = e^x \log x + C \]The problem specifies that the integral equals \(e^x f(x) + C\).
By comparing \(e^x \log x + C\) with \(e^x f(x) + C\), we deduce:\[ f(x) = \log x \]Step 4: Final Result:
The function f(x) is \(\log x\).