Indefinite Integrals
Evaluate ∫ex(x1+xlnx)dx (Assume x>0 and base e for ln).
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About This Question
- Subject
- mathematics
- Chapter
- integral calculus
- Topic
- indefinite integrals
- Difficulty
- Easy
- Year
- 2025
Solution
Correct Answer:
exlnx+C
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Simplify the integrand: x1+xlnx=x1+lnx.
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Recognize the standard form ∫ex(f(x)+f′(x))dx=exf(x). Let f(x)=lnx. Then f′(x)=1/x.
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The integral becomes: ∫ex(lnx+x1)dx=exlnx+C.
This easy difficulty mathematics question is from the chapter integral calculus, covering the topic of indefinite integrals. It appeared in the 2025 exam.
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