∫dx/(√(5-4x-x^2)) is equal to

The correct answer is Option (1) → \(\sin^{-1}\left(\displaystyle\frac{x+2}{3}\right)+C\): C is an arbitrary constant

Completing the square under the root:

    \(5 - 4x - x^2 = 5 - (x^2 + 4x)\)
    \(= 5 - [(x + 2)^2 - 4]\)
    \(= 9 - (x + 2)^2\)

So

    \(\displaystyle \int\frac{dx}{\sqrt{5 - 4x - x^2}} = \int\frac{dx}{\sqrt{3^2 - (x + 2)^2}}\)

Using the standard integral

    \(\displaystyle \int\frac{dx}{\sqrt{a^2 - u^2}} = \sin^{-1}\frac{u}{a} + C\)

with u = x + 2 and a = 3:

    \(\displaystyle = \sin^{-1}\left(\frac{x + 2}{3}\right) + C\)

Tip: \(\int\frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\frac{x}{a} + C\) has no factor \(\frac{1}{a}\), unlike \(\int\frac{dx}{a^2 + x^2} = \frac{1}{a}\tan^{-1}\frac{x}{a} + C\). Mixing the two gives Option (2).