$\int \sqrt{x^2-a^2}\,dx$ equals

Question

Accepted Answer

B. $\dfrac{x}{2}\sqrt{x^2-a^2}-\dfrac{a^2}{2}\log\left|x+\sqrt{x^2-a^2}\right|+C$

1. Take constant 1 as second function and integrate by parts.
2. This yields $2I=x\sqrt{x^2-a^2}-a^2\int\dfrac{dx}{\sqrt{x^2-a^2}}$.
3. Using $\int\dfrac{dx}{\sqrt{x^2-a^2}}=\log|x+\sqrt{x^2-a^2}|$.
4. Result: $\dfrac{x}{2}\sqrt{x^2-a^2}-\dfrac{a^2}{2}\log|x+\sqrt{x^2-a^2}|+C$. Option A has the wrong sign; C is the $a^2-x^2$ case.
_Source: NCERT Class 12 Mathematics Ch 7 "Integrals", p.330_

Answer

$\dfrac{x}{2}\sqrt{x^2-a^2}+\dfrac{a^2}{2}\log\left|x+\sqrt{x^2-a^2}\right|+C$

Answer

$\dfrac{x}{2}\sqrt{x^2-a^2}+\dfrac{a^2}{2}\sin^{-1}\dfrac{x}{a}+C$

Answer

$\dfrac{1}{3}(x^2-a^2)^{3/2}+C$