allery$\int \dfrac{1}{x}\,dx$ equals
A$-\dfrac{1}{x^2}+C$
B$x\log|x|+C$
C$\dfrac{x^2}{2}+C$
D$\log|x|+C$
Answer & Solution
Correct answer: D. $\log|x|+C$
1. The power rule fails for $n=-1$.
2. Recall $\dfrac{d}{dx}(\log|x|)=\dfrac{1}{x}$ for $x\neq 0$.
3. Hence $\int \dfrac{1}{x}\,dx=\log|x|+C$.
_Source: syllabus Mathematics Ch 7 "Integrals", p.290_
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