Let's break this down step by step.

Given Integral:

$\int \frac{x + 5}{(x + 2)(x^2 + 3)} \, dx$

We are tasked with applying the method of partial fractions to express the integrand as a sum of simpler rational functions.

Step 1: Set up the partial fraction decomposition

We want to express the rational function $\frac{x + 5}{(x + 2)(x^2 + 3)}$ as a sum of simpler fractions. To do this, we write:

$\frac{x + 5}{(x + 2)(x^2 + 3)} = \frac{A}{x + 2} + \frac{Bx + C}{x^2 + 3}$

Where $A$, $B$, and $C$ are constants to be determined.

Step 2: Multiply both sides by $(x + 2)(x^2 + 3)$

To eliminate the denominators, we multiply both sides by $(x + 2)(x^2 + 3)$:

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

Now expand both sides.

• On the left-hand side, we have: $x + 5$

• On the right-hand side, expand each term: $A(x^2 + 3) = A x^2 + 3A$ $(Bx + C)(x + 2) = Bx^2 + 2Bx + Cx + 2C$ Combine the terms on the right-hand side: $A x^2 + 3A + B x^2 + 2Bx + Cx + 2C$ Group like terms: $(A + B) x^2 + (2B + C) x + (3A + 2C)$

Step 3: Equate coefficients of like powers of $x$

Now we match the coefficients from both sides of the equation:

• Coefficient of $x^2$: $A + B = 0$
• Coefficient of $x$: $2B + C = 1$
• Constant term: $3A + 2C = 5$

Step 4: Solve the system of equations

We now solve the system of equations:

1. $A + B = 0$ ⟹ $A = -B$
2. $2B + C = 1$
3. $3A + 2C = 5$

Substitute $A = -B$ into the second and third equations:

• From $2B + C = 1$, we get $C = 1 - 2B$.
• Substitute $C = 1 - 2B$ into $3A + 2C = 5$: $3(-B) + 2(1 - 2B) = 5$ Simplify: $-3B + 2 - 4B = 5$ $-7B + 2 = 5$ $-7B = 3$ $B = -\frac{3}{7}$

Now substitute $B = -\frac{3}{7}$ into $A = -B$ and $C = 1 - 2B$:

• $A = \frac{3}{7}$
• $C = 1 - 2(-\frac{3}{7}) = 1 + \frac{6}{7} = \frac{13}{7}$

Step 5: The partial fraction decomposition

Thus, the partial fraction decomposition is:

$\frac{x + 5}{(x + 2)(x^2 + 3)} = \frac{3/7}{x + 2} + \frac{-3x/7 + 13/7}{x^2 + 3}$

Step 6: Antiderivative

Now, we can find the antiderivative of each term:

1. $\int \frac{3/7}{x + 2} \, dx = \frac{3}{7} \ln|x + 2|$
2. For $\int \frac{-3x/7 + 13/7}{x^2 + 3} \, dx$, we split it into two parts:
   • $\int \frac{-3x/7}{x^2 + 3} \, dx = -\frac{3}{14} \ln(x^2 + 3)$
   • $\int \frac{13/7}{x^2 + 3} \, dx = \frac{13}{7\sqrt{3}} \arctan\left(\frac{x}{\sqrt{3}}\right)$

Thus, the antiderivative is:

$\frac{3}{7} \ln|x + 2| - \frac{3}{14} \ln(x^2 + 3) + \frac{13}{7\sqrt{3}} \arctan\left(\frac{x}{\sqrt{3}}\right) + C$

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Would you like me to elaborate on any of the steps or answer any questions you may have? Here are some related questions to explore further:

1. How does the method of partial fractions work for other types of rational functions?
2. Can you explain how the arctangent function arises when integrating rational functions with quadratic denominators?
3. What are the conditions under which partial fraction decomposition can be applied?
4. How would you perform partial fraction decomposition for higher-degree polynomials in the numerator or denominator?
5. How do we handle cases where the denominator has repeated factors in partial fraction decomposition?

Tip: When performing partial fraction decomposition, always ensure the degree of the numerator is less than the degree of the denominator. If not, perform polynomial long division first!