Find the integral: $\int \frac{x+3}{\sqrt{5-4x-x^2}} dx$

$-\sqrt{5-4x-x^2} + \sin^{-1}(\frac{x+2}{3}) + C$

The correct answer is Option (1) → $-\sqrt{5-4x-x^2} + \sin^{-1}(\frac{x+2}{3}) + C$

Let us express

$x + 3 = A \frac{d}{dx} (5 - 4x - x^2) + B = A(-4 - 2x) + B$

Equating the coefficients of $x$ and the constant terms from both sides, we get

$-2A = 1$ and $-4A + B = 3$, i.e., $A = -\frac{1}{2}$ and $B = 1$

Therefore, $\int \frac{x+3}{\sqrt{5-4x-x^2}} \, dx = -\frac{1}{2} \int \frac{(-4-2x) \, dx}{\sqrt{5-4x-x^2}} + \int \frac{dx}{\sqrt{5-4x-x^2}}$

$= -\frac{1}{2} I_1 + I_2 \dots (1)$

In $I_1$, put $5 - 4x - x^2 = t$, so that $(-4-2x) \, dx = dt$.

Therefore,

$I_1 = \int \frac{(-4-2x) \, dx}{\sqrt{5-4x-x^2}} = \int \frac{dt}{\sqrt{t}} = 2\sqrt{t} + C_1$

$= 2\sqrt{5-4x-x^2} + C_1 \dots (2)$

Now consider

$I_2 = \int \frac{dx}{\sqrt{5-4x-x^2}} = \int \frac{dx}{\sqrt{9-(x+2)^2}}$

Put $x + 2 = t$, so that $dx = dt$.

Therefore,

$I_2 = \int \frac{dt}{\sqrt{3^2 - t^2}} = \sin^{-1} \frac{t}{3} + C_2$

$= \sin^{-1} \frac{x+2}{3} + C_2 \dots (3)$

Substituting (2) and (3) in (1), we obtain

$\int \frac{x+3}{\sqrt{5-4x-x^2}} = -\sqrt{5-4x-x^2} + \sin^{-1} \frac{x+2}{3} + C \text{, where } C = C_2 - \frac{C_1}{2}$