allery$\int \dfrac{dx}{x^2+a^2}$ equals
A$\dfrac{1}{2a}\log\left|\dfrac{x-a}{x+a}\right|+C$
B$\dfrac{1}{a}\tan^{-1}\dfrac{x}{a}+C$
C$\sin^{-1}\dfrac{x}{a}+C$
D$\log\left|x+\sqrt{x^2+a^2}\right|+C$
Answer & Solution
Correct answer: B. $\dfrac{1}{a}\tan^{-1}\dfrac{x}{a}+C$
1. Substitute $x=a\tan\theta$, so $dx=a\sec^2\theta\,d\theta$.
2. Denominator $=a^2\tan^2\theta+a^2=a^2\sec^2\theta$.
3. Integral $=\dfrac{1}{a}\int d\theta=\dfrac{\theta}{a}=\dfrac{1}{a}\tan^{-1}\dfrac{x}{a}+C$.
4. Option A is for $\dfrac{1}{x^2-a^2}$.
_Source: syllabus Mathematics Ch 7 "Integrals", p.307_
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