$\int \sqrt{3-2x-x^2}\,dx$ equals

Question

Accepted Answer

C. $\dfrac{1}{2}(x+1)\sqrt{3-2x-x^2}+2\sin^{-1}\!\left(\dfrac{x+1}{2}\right)+C$

1. Complete the square: $3-2x-x^2=4-(x+1)^2$.
2. Put $y=x+1$: integral $=\int \sqrt{4-y^2}\,dy$.
3. Use $\int \sqrt{a^2-y^2}\,dy=\dfrac{y}{2}\sqrt{a^2-y^2}+\dfrac{a^2}{2}\sin^{-1}\dfrac{y}{a}$ with $a=2$.
4. Back-substitute: option C. Option A wrongly uses a log (that is the $x^2+a^2$ form).
_Source: NCERT Class 12 Mathematics Ch 7 "Integrals", p.331_

Answer

$\dfrac{1}{2}(x+1)\sqrt{3-2x-x^2}+2\log\left|x+1+\sqrt{3-2x-x^2}\right|+C$

Answer

$\dfrac{1}{2}(x+1)\sqrt{3-2x-x^2}-2\sin^{-1}\!\left(\dfrac{x+1}{2}\right)+C$

Answer

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