The expression in the image is:

$\frac{x + 3}{(x^2 + 4)(x + 5)}$

To resolve this into partial fractions, we express the given fraction as a sum of simpler fractions:

$\frac{x + 3}{(x^2 + 4)(x + 5)} = \frac{Ax + B}{x^2 + 4} + \frac{C}{x + 5}$

Here, $\frac{Ax + B}{x^2 + 4}$ corresponds to the irreducible quadratic factor, and $\frac{C}{x + 5}$ corresponds to the linear factor.

Step 1: Combine the fractions on the right-hand side

To combine the fractions, we get a common denominator:

$\frac{Ax + B}{x^2 + 4} + \frac{C}{x + 5} = \frac{(Ax + B)(x + 5) + C(x^2 + 4)}{(x^2 + 4)(x + 5)}$

Step 2: Set the numerators equal

Since the denominators are the same, we equate the numerators:

$x + 3 = (Ax + B)(x + 5) + C(x^2 + 4)$

Step 3: Expand and collect like terms

Expanding the right-hand side:

$(Ax + B)(x + 5) = Ax^2 + 5Ax + Bx + 5B = Ax^2 + (5A + B)x + 5B$ $C(x^2 + 4) = Cx^2 + 4C$

Combining these:

$Ax^2 + (5A + B)x + 5B + Cx^2 + 4C = (A + C)x^2 + (5A + B)x + (5B + 4C)$

Step 4: Match coefficients

We now match the coefficients of like terms from both sides of the equation:

For $x^2$ terms: $A + C = 0$

For $x$ terms: $5A + B = 1$

For the constant term: $5B + 4C = 3$

Step 5: Solve the system of equations

From $A + C = 0$, we get: $C = -A$

Substitute $C = -A$ into the other equations:

For $5A + B = 1$: $B = 1 - 5A$

For $5B + 4C = 3$: $5(1 - 5A) + 4(-A) = 3$ $5 - 25A - 4A = 3$ $-29A = -2$ $A = \frac{2}{29}$

Now, substitute $A = \frac{2}{29}$ back to find $B$ and $C$:

$B = 1 - 5\left(\frac{2}{29}\right) = 1 - \frac{10}{29} = \frac{29}{29} - \frac{10}{29} = \frac{19}{29}$ $C = -A = -\frac{2}{29}$

Step 6: Write the partial fraction decomposition

Thus, the partial fraction decomposition is:

$\frac{x + 3}{(x^2 + 4)(x + 5)} = \frac{\frac{2}{29}x + \frac{19}{29}}{x^2 + 4} + \frac{-\frac{2}{29}}{x + 5}$

Would you like more details or have any questions?

Here are some related questions you might find interesting:

1. How do you decompose a fraction with repeated quadratic factors?
2. Can you explain the method of partial fractions when the denominator has repeated linear factors?
3. How does the degree of the numerator affect the decomposition in partial fractions?
4. What happens when the numerator’s degree is higher than the denominator's in partial fractions?
5. How can partial fraction decomposition be used in solving integrals?

Tip: When solving partial fractions, always check that the degree of the numerator is less than the degree of the denominator before beginning the decomposition.