Key Concepts and Formulas

• Completing the Square: $ax^2 + bx + c = a(x + \frac{b}{2a})^2 + (c - \frac{b^2}{4a})$
• Integration of $\frac{f'(x)}{\sqrt{f(x)}}$: $\int \frac{f'(x)}{\sqrt{f(x)}} dx = 2\sqrt{f(x)} + C$
• Integration of $\frac{1}{\sqrt{a^2 - x^2}}$: $\int \frac{1}{\sqrt{a^2 - x^2}} dx = \sin^{-1}(\frac{x}{a}) + C$

Step-by-Step Solution

Step 1: Rewrite the integral to make it easier to integrate. We start with the given integral: $\int {{{2x + 5} \over {\sqrt {7 - 6x - {x^2}} }}} \,\,dx$ Our goal is to manipulate the numerator to resemble the derivative of the expression inside the square root, which is $7 - 6x - x^2$. The derivative is $-(6 + 2x) = -2(x+3) = -2x - 6$. We rewrite the numerator $2x + 5$ as $(2x + 6) - 1$. This is done to separate the integral into two parts, one of which can be solved directly using the derivative of the inside function, and the other using a standard integral formula. $\int {{{2x + 5} \over {\sqrt {7 - 6x - {x^2}} }}} dx = \int {{{(2x + 6) - 1} \over {\sqrt {7 - 6x - {x^2}} }}} dx$

Step 2: Split the integral into two separate integrals. $\int {{{(2x + 6) - 1} \over {\sqrt {7 - 6x - {x^2}} }}} dx = \int {{{2x + 6} \over {\sqrt {7 - 6x - {x^2}} }}} dx - \int {{1 \over {\sqrt {7 - 6x - {x^2}} }}} dx$

Step 3: Evaluate the first integral. Let $I_1 = \int {{{2x + 6} \over {\sqrt {7 - 6x - {x^2}} }}} dx$. We notice that the numerator is almost the negative of the derivative of the expression inside the square root. More precisely, $\frac{d}{dx}(7 - 6x - x^2) = -6 - 2x = -(2x+6)$. Therefore, we can rewrite the integral as: $I_1 = -\int {{\frac{-2x - 6}{\sqrt {7 - 6x - {x^2}} }}} dx = - \int {{\frac{-(2x + 6)}{\sqrt {7 - 6x - {x^2}} }}} dx$ Using the formula $\int \frac{f'(x)}{\sqrt{f(x)}} dx = 2\sqrt{f(x)} + C$, we have: $I_1 = -2\sqrt{7 - 6x - x^2} + C_1$

Step 4: Evaluate the second integral. Let $I_2 = \int {{1 \over {\sqrt {7 - 6x - {x^2}} }}} dx$. We need to complete the square for the expression inside the square root: $7 - 6x - x^2$. $7 - 6x - x^2 = -(x^2 + 6x) + 7 = -(x^2 + 6x + 9 - 9) + 7 = -(x+3)^2 + 9 + 7 = 16 - (x+3)^2 = 4^2 - (x+3)^2$ Thus, $I_2 = \int {{1 \over {\sqrt {4^2 - (x + 3)^2} }}} dx$ Using the formula $\int \frac{1}{\sqrt{a^2 - x^2}} dx = \sin^{-1}(\frac{x}{a}) + C$, we have: $I_2 = \sin^{-1}\left(\frac{x+3}{4}\right) + C_2$

Step 5: Combine the results from Step 3 and Step 4. $\int {{{2x + 5} \over {\sqrt {7 - 6x - {x^2}} }}} dx = I_1 - I_2 = -2\sqrt{7 - 6x - x^2} - \sin^{-1}\left(\frac{x+3}{4}\right) + C$ where $C = C_1 - C_2$.

Step 6: Compare with the given equation. We are given that: $\int {{{2x + 5} \over {\sqrt {7 - 6x - {x^2}} }}} \,\,dx = A\sqrt {7 - 6x - {x^2}} + B{\sin ^{ - 1}}\left( {{{x + 3} \over 4}} \right) + C$ Comparing this with our result: $-2\sqrt{7 - 6x - x^2} - \sin^{-1}\left(\frac{x+3}{4}\right) + C$ We can see that $A = -2$ and $B = -1$.

Therefore, the ordered pair (A, B) is (-2, -1).

Common Mistakes & Tips

• Sign Errors: Be very careful with the signs when completing the square and when taking derivatives.
• Completing the Square Correctly: Ensure you complete the square accurately. An error here will propagate through the entire solution.
• Recognizing Standard Integrals: Practice recognizing standard integral forms so you can apply the formulas directly.

Summary

We started by rewriting the integrand to separate it into two parts: one that could be solved using the derivative of the expression inside the square root and another that could be solved using a standard inverse sine integral. By completing the square, we were able to transform the second integral into a recognizable form. Finally, by comparing our result with the given equation, we determined the values of A and B to be -2 and -1, respectively.

Final Answer The final answer is \boxed{(-2, -1)}, which corresponds to option (B).