allery$\int \dfrac{dx}{x^2-16}$ equals
A$\dfrac{1}{4}\tan^{-1}\dfrac{x}{4}+C$
B$\dfrac{1}{8}\log\left|\dfrac{x-4}{x+4}\right|+C$
C$\dfrac{1}{8}\log\left|\dfrac{x+4}{x-4}\right|+C$
D$\sin^{-1}\dfrac{x}{4}+C$
Answer & Solution
Correct answer: B. $\dfrac{1}{8}\log\left|\dfrac{x-4}{x+4}\right|+C$
1. Here $x^2-16=x^2-4^2$, so $a=4$.
2. Use $\int \dfrac{dx}{x^2-a^2}=\dfrac{1}{2a}\log\left|\dfrac{x-a}{x+a}\right|+C$.
3. With $2a=8$: $\dfrac{1}{8}\log\left|\dfrac{x-4}{x+4}\right|+C$.
4. Option C inverts the fraction (wrong sign).
_Source: syllabus Mathematics Ch 7 "Integrals", p.310_
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