Question:
$\int_1^e \frac{e^x}{x}(1 + x \log x) dx =$
$\int_1^e \frac{e^x}{x}(1 + x \log x) dx =$
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Look for derivative of product.
Updated On: Apr 26, 2026
- $e^e$
- $e^e - e$
- $e^e + e$
- $e$
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The Correct Option is A
Solution and Explanation
Concept:
Recognize derivative pattern. Step 1: Observe. \[ \frac{d}{dx}(e^x \log x) = \frac{e^x}{x}(1+x\log x) \]
Step 2: Integrate. \[ = e^x \log x \Big|_1^e \]
Step 3: Evaluate. \[ = e^e - 0 = e^e \] Conclusion. Answer = $e^e$
Recognize derivative pattern. Step 1: Observe. \[ \frac{d}{dx}(e^x \log x) = \frac{e^x}{x}(1+x\log x) \]
Step 2: Integrate. \[ = e^x \log x \Big|_1^e \]
Step 3: Evaluate. \[ = e^e - 0 = e^e \] Conclusion. Answer = $e^e$
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