Problem 38 - College Algebra

Question

Write the partial fraction decomposition of each rational expression. $$\frac{x^{2}+2 x+3}{\left(x^{2}+4\right)^{2}}$$

Step-by-Step Solution

Answer

The partial fraction decomposition of the rational expression \(\frac{x^{2}+2 x+3}{\left(x^{2}+4\right)^{2}}\) is \(\frac{x+3}{(x^2+4)^2}\).

Step 1: Identify the Form of the Partial Fractions

Given the rational expression \(\frac{x^{2}+2 x+3}{\left(x^{2}+4\right)^{2}}\), it can be seen that the denominator is a square of a binomial (that is, in the form \( (x^2+n)^2 \)). Hence, the partial fraction decomposition will have the forms: \( \frac{A}{{x^2+4}} + \frac{Bx+C}{{(x^2+4)}^2} \).

Step 2: Setting up Equations

The above forms can be rewritten as a single fraction: \[ \frac{A}{{x^2+4}} + \frac{Bx+C}{{(x^2+4)}^2} = \frac{x^{2}+2 x+3}{\left(x^{2}+4\right)^{2}} \] This gives: \(Ax (x^2+4) + (Bx+C) = x^{2}+2 x+3\). By comparing the coefficients on both sides, equations can be formed for A, B, and C.

Step 3: Solving for A, B, and C

From the coefficient of \(x^3\), A should be equal to 0, because there is no \(x^3\) term on the right side. For the coefficient of \(x^2\), 4A + B should be 1, as the coefficient of \(x^2\) on the right side is 1. But since we have A = 0, we get B = 1. The equation for C resulting from the constant term is 4A+C = 3, which with A = 0, gives C = 3. With these values, the partial fraction decomposition is: \(\frac{0}{x^2+4} + \frac{x+3}{(x^2+4)^2}\), simplified to \(\frac{x+3}{(x^2+4)^2}\).