allery$\displaystyle\int \dfrac{1}{x^2 - a^2}\,dx$ (with $|x| > a > 0$) equals:
A$\dfrac{1}{2a}\log\left|\dfrac{x-a}{x+a}\right| + C$
B$\dfrac{1}{2a}\log\left|\dfrac{x+a}{x-a}\right| + C$
C$\dfrac{1}{a}\log|x^2 - a^2| + C$
D$\dfrac{1}{a}\tan^{-1}\dfrac{x}{a} + C$
Answer & Solution
Correct answer: A. $\dfrac{1}{2a}\log\left|\dfrac{x-a}{x+a}\right| + C$
**Decompose:** $\dfrac{1}{x^2 - a^2} = \dfrac{1}{(x-a)(x+a)} = \dfrac{1}{2a}\left(\dfrac{1}{x-a} - \dfrac{1}{x+a}\right)$.
**Integrating:** $\dfrac{1}{2a}\bigl(\log|x-a| - \log|x+a|\bigr) + C = \dfrac{1}{2a}\log\left|\dfrac{x-a}{x+a}\right| + C$.
Keep practising Unit 6A: Indefinite Integrals & Fundamental Theorem
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