Calculus

Integrals

Medium

core

$\int \frac{dx}{\sqrt{5 - 4x - x^2}}$ 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

✅ Correct Option: 1

The integral $\int \frac{dx}{\sqrt{5 - 4x - x^2}}$ requires converting the expression under the square root into standard form.

The standard formula needed is:

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

To convert $5 - 4x - x^2$ into the form $a^2 - u^2$ , complete the square.

Taking out the negative sign from the $x^2$ and $x$ terms:

$5 - 4x - x^2 = -(x^2 + 4x) + 5$

Adding and subtracting $\left(\frac{4}{2}\right)^2 = 4$ inside the bracket:

$= -(x^2 + 4x + 4 - 4) + 5$

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

$= -(x+2)^2 + 4 + 5$

$= 9 - (x+2)^2$

$= 3^2 - (x+2)^2$

The integral becomes:

$\int \frac{dx}{\sqrt{9 - (x+2)^2}}$

This matches the standard form with $a = 3$ and $u = (x+2)$ .

Applying the standard formula:

$\int \frac{dx}{\sqrt{9 - (x+2)^2}} = \sin^{-1}\left(\frac{x+2}{3}\right) + C$

Therefore, the answer is $\sin^{-1}\left(\frac{x + 2}{3}\right) + C$

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∫dx5−4x−x2\int \frac{dx}{\sqrt{5 - 4x - x^2}}∫5−4x−x2​dx​ is equal to