Thank you for the clarification. The integral is:

$\int (x + 3) \sqrt{3 - 4x - x^2} \, dx$

Let's solve this step-by-step using substitution.

We will use the substitution:

$u = 3 - 4x - x^2$

First, differentiate $u$ with respect to $x$:

$\frac{du}{dx} = -4 - 2x$

Thus,

$du = (-4 - 2x) \, dx$

Solve for $dx$:

$dx = \frac{du}{-4 - 2x}$

Now, we need to express $(x + 3) \sqrt{3 - 4x - x^2}$ in terms of $u$.

Notice that:

$x + 3$

is already part of our substitution context. Express $x$ in terms of $u$:

$x = -2 \pm \sqrt{-1 - u}$

However, simpler is to handle $x + 3$ combined:

Next integral becomes:

$I = \int \frac{(x + 3)\sqrt{u} \, du}{-2(x + 2)}$

Splitting into manageable parts directly substituting often requires adjusting constants:

$\int u^{\frac{1}{2}} du = \int u^{\frac{1}{2}} \cdot \frac{du}{-2x-4}$

Splitting constants simplifies:

$I = \int -1/2 u^{1/2} \, du$

Solving directly:

$\int -1/2 u^{1/2} du = -1/3 u^{3/2} + C$

Back-substitute $u = 3 - 4x - x^2$:

$\int (x+3)\sqrt{3-4x-x^2} \, dx = -1/3 (3-4x-x^2)^{3/2} + C$

This represents the integral.

Would you like further details or explore related methods?

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Tip: Always re-check substitution validity and constants throughout integration process ensuring clear substitutions.