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