To solve for $A$, $B$, and $C$ in the equation:

$\frac{x^2 + x + 3}{(x^2 + 5)(x - 3)} = \frac{Ax + B}{x^2 + 5} + \frac{C}{x - 3}$

we can use partial fraction decomposition. Here's the step-by-step approach:

1. Set up the equation: Multiply both sides by the common denominator $(x^2 + 5)(x - 3)$ to eliminate the fractions:

   $x^2 + x + 3 = (Ax + B)(x - 3) + C(x^2 + 5)$

2. Expand the terms: Expand the right-hand side:

   $x^2 + x + 3 = (Ax + B)(x - 3) + Cx^2 + 5C$

   Expanding $(Ax + B)(x - 3)$ gives:

   $Ax^2 - 3Ax + Bx - 3B$

   So the equation becomes:

   $x^2 + x + 3 = Ax^2 - 3Ax + Bx - 3B + Cx^2 + 5C$

3. Combine like terms: Group terms by powers of $x$:

   • The $x^2$ terms: $(A + C)x^2$
   • The $x$ terms: $(-3A + B)x$
   • The constant terms: $-3B + 5C$

   Now we rewrite the equation:

   $x^2 + x + 3 = (A + C)x^2 + (-3A + B)x + (-3B + 5C)$

4. Set up a system of equations: By equating the coefficients on both sides, we get:

   • For $x^2$: $A + C = 1$
   • For $x$: $-3A + B = 1$
   • For the constant term: $-3B + 5C = 3$

5. Solve the system of equations:

   From $A + C = 1$: $C = 1 - A$

   Substitute $C = 1 - A$ into the other equations.

   • Substituting into $-3A + B = 1$: $B = 1 + 3A$

   • Substituting into $-3B + 5C = 3$: $-3(1 + 3A) + 5(1 - A) = 3$

     Simplifying this:

     $-3 - 9A + 5 - 5A = 3$

     $-14A + 2 = 3$

     $-14A = 1$

     $A = -\frac{1}{14}$

   Now substitute $A = -\frac{1}{14}$ back to find $B$ and $C$:

   • For $B$: $B = 1 + 3\left(-\frac{1}{14}\right) = 1 - \frac{3}{14} = \frac{14 - 3}{14} = \frac{11}{14}$

   • For $C$: $C = 1 - \left(-\frac{1}{14}\right) = 1 + \frac{1}{14} = \frac{14 + 1}{14} = \frac{15}{14}$

6. Solution: The values are:

   $A = -\frac{1}{14}, \quad B = \frac{11}{14}, \quad C = \frac{15}{14}$

Let me know if you would like further details or have any questions.

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1. How can partial fraction decomposition help in solving integrals?
2. What happens if we factor a higher power in the denominator?
3. How do these solutions change if we change the terms in the denominator?
4. Why is partial fraction decomposition useful in calculus?
5. Can you apply a similar method to solve equations with complex roots?

Tip: Always check the system of equations carefully after substitution to avoid errors!