$\int \dfrac{dx}{x^2-a^2}$ equals
A$\dfrac{1}{a}\tan^{-1}\dfrac{x}{a}+C$
B$\dfrac{1}{2a}\log\left|\dfrac{a+x}{a-x}\right|+C$
C$\dfrac{1}{2a}\log\left|\dfrac{x-a}{x+a}\right|+C$
D$\log\left|x+\sqrt{x^2-a^2}\right|+C$
Answer & Solution
Correct answer: C. $\dfrac{1}{2a}\log\left|\dfrac{x-a}{x+a}\right|+C$
1. Factor: $\dfrac{1}{(x-a)(x+a)}=\dfrac{1}{2a}\left[\dfrac{1}{x-a}-\dfrac{1}{x+a}\right]$.
2. Integrate term by term: $\dfrac{1}{2a}[\log|x-a|-\log|x+a|]$.
3. Combine: $\dfrac{1}{2a}\log\left|\dfrac{x-a}{x+a}\right|+C$.
4. Option B is for $\dfrac{1}{a^2-x^2}$ (signs reversed).
_Source: NCERT Class 12 Mathematics Ch 7 "Integrals", p.306_
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