$\int\frac{dx}{\sqrt{5-4x-x^2}}$ is equal to
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → $\sin^{-1}(\frac{x+2}{3})+C$: C is an arbitrary constant
$I=\int \frac{dx}{\sqrt{5-4x-x^{2}}}$
$5-4x-x^{2}=9-(x+2)^{2}$
$I=\int \frac{dx}{\sqrt{9-(x+2)^{2}}}$
$\int \frac{dx}{\sqrt{a^{2}-(x-c)^{2}}}=\sin^{-1}\!\left(\frac{x-c}{a}\right)+C$
$\Rightarrow I=\sin^{-1}\!\left(\frac{x+2}{3}\right)+C$
$\sin^{-1}\!\left(\frac{x+2}{3}\right)+C$