Chapter 11.5, Problem 42AYU

To determine

To find: The partial fraction decomposition of the given rational function.

Answer to Problem 42AYU

Solution:

$\frac{1}{{\left(x^2 + 4\right)}^2} - \frac{4}{{\left(x^2 + 4\right)}^3}$

Explanation of Solution

Given:

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

Formula used:

a. If $Q\left(x\right)$ contains a non-repeated irreducible quadratic factor of the form $a{x}^2 + bx + c$, then, in the partial fraction decomposition of $P\left(x\right)/Q\left(x\right)$, takes the form:

$\frac{Ax + B}{a{x}^2 + bx + c}$

where the numbers A and B are to be determined.

b. If $Q$ has repeated linear factor of the form ${\left(x - a\right)}^n, \text{n} \ge \text{2}$ an integer, then, in the partial fraction decomposition of $P\left(x\right)/Q\left(x\right)$, takes the form:

$\frac{P\left(x\right)}{Q\left(x\right)} = \frac{A_1}{\left(x - a\right)} + \frac{A_2}{{\left(x - a\right)}^2} + \cdots + \frac{A_n}{{\left(x - a\right)}^n}$

where the numbers $A_1, A_2\dots A_n$ are to be determined.

Calculation:

The denominator has repeated linear irreducible quadratic factor. Therefore the partial fraction decomposition takes the form

$\frac{x^2}{{\left(x^2 + 4\right)}^3} = \frac{Ax + B}{\left(x^2 + 4\right)} + \frac{Cx + D}{{\left(x^2 + 4\right)}^2} + \frac{Ex + F}{{\left(x^2 + 4\right)}^3}$

$\begin{array}{ll}\Rightarrow x^4\hfill & = \left(Ax + B\right){\left(x^2 + 4\right)}^2 + \left(Cx + D\right)\left(x^2 + 4\right) + Ex + F\hfill \\ \hfill & = \left(Ax + B\right)\left(x^4 + 16 + 8x^2\right) + \left(Cx + D\right)\left(x^2 + 4\right) + Ex + F\hfill \end{array}$   -----Eq (1)

$x^5, \Rightarrow A = 0$   -----Eq(2)

$x^4, \Rightarrow B = 0$   -----Eq(3)

$x^3 \Rightarrow 8A + C = 0$   -----Eq(4)

$x^2, \Rightarrow 8B + D = 1$   -----Eq(5)

$x, \Rightarrow 16A + 4C + E = 0$   -----Eq(6)

$constant \Rightarrow 16B + 4D + F = 0$   -----Eq(7)

Solving for $A,B, C,E,F$ from Eq(2) – Eq(7)we get

$A = 0, B = 0, C = 0, D = 1, E = 0 and F = \text{−}4$

The partial fraction decomposition of $\frac{x^2}{{\left(x^2 + 4\right)}^3}$ is $\frac{1}{{\left(x^2 + 4\right)}^2} - \frac{4}{{\left(x^2 + 4\right)}^3}$