$\int \dfrac{1}{x}\,dx$ equals
- $-\dfrac{1}{x^2}+C$
- $x\log|x|+C$
- $\dfrac{x^2}{2}+C$
- $\log|x|+C$
Show answer and 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_