The problem from the image is asking for the partial fraction decomposition of the expression:

$\frac{5x + 6}{x(x + 3)^2}$

Solution:

To decompose this into partial fractions, we start by writing it in the form:

$\frac{5x + 6}{x(x+3)^2} = \frac{A}{x} + \frac{B}{x+3} + \frac{C}{(x+3)^2}$

Now, multiply both sides by the denominator $x(x+3)^2$ to eliminate the fractions:

$5x + 6 = A(x+3)^2 + Bx(x+3) + Cx$

Expand both sides:

$5x + 6 = A(x^2 + 6x + 9) + Bx(x + 3) + Cx$

$5x + 6 = A(x^2 + 6x + 9) + B(x^2 + 3x) + Cx$

Now expand each term:

$5x + 6 = A(x^2) + A(6x) + A(9) + B(x^2) + B(3x) + C(x)$

$5x + 6 = (A + B)x^2 + (6A + 3B + C)x + 9A$

Now, compare the coefficients of $x^2$, $x$, and the constant terms on both sides:

• For $x^2$: $A + B = 0$
• For $x$: $6A + 3B + C = 5$
• For the constant: $9A = 6$

Solving the system of equations:

1. $9A = 6 \Rightarrow A = \frac{2}{3}$
2. From $A + B = 0 \Rightarrow B = -A = -\frac{2}{3}$
3. Substitute $A = \frac{2}{3}$ and $B = -\frac{2}{3}$ into $6A + 3B + C = 5$:

$6\left(\frac{2}{3}\right) + 3\left(-\frac{2}{3}\right) + C = 5$

$4 - 2 + C = 5 \Rightarrow C = 3$

Final Partial Fraction Decomposition:

$\frac{5x + 6}{x(x+3)^2} = \frac{2/3}{x} - \frac{2/3}{x+3} + \frac{3}{(x+3)^2}$

Or, you can rewrite it as:

$\frac{5x + 6}{x(x+3)^2} = \frac{2}{3x} - \frac{2}{3(x+3)} + \frac{3}{(x+3)^2}$

Would you like more details on any step, or have any questions?

Related questions:

1. How do you approach partial fraction decomposition when the denominator has higher powers?
2. What happens if the numerator's degree is higher than the denominator in partial fractions?
3. How can partial fractions be used in integral calculus?
4. What are the conditions for a unique partial fraction decomposition?
5. How would the decomposition change if there were irreducible quadratic factors in the denominator?

Tip: Always simplify or factor the denominator fully before attempting partial fraction decomposition to ensure the correct form of fractions.