Chapter 11.5, Problem 27AYU

To determine

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

Answer to Problem 27AYU

Solution:

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

Explanation of Solution

Given:

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

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 $\text{A}$ and $\text{B}$ are to be determined.

b. If $Q$ has repeated linear factor of the form ${\left(x - a\right)}^n, \text{n} \ge 2$, 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 $A_{1, }{A}_2\mathrm{...}{A}_n$ are to be determined.

Calculation:

The given partial fraction decomposition takes the form

$\frac{x + 4}{x^2\left(x^2 + 4\right)} = = \frac{A}{x} + \frac{B}{x^2} + \frac{cx + D}{\left(x^2 + 4\right)}$

$= >x + 4 = Ax\left(x^2 + 4\right) + B\left(x^2 + 4\right) + \left(Cx + D\right)x^2$    -----Eq(1)

$\begin{array}{cc}& = A\end{array}\left(x^3 + 4x\right) + B\left(x^2 + 4\right) + C{x}^3 + D{x}^2$

Equating the co-efficients on both sides in Eq(1),

$\begin{array}{cc}{x}^3, & = >A + C = 0\end{array}$   -----Eq(2)

$\begin{array}{cc}{x}^2& = >B + D = 0\end{array}$   -----Eq(3)

$\begin{array}{cc}x, & = >4A = 1\end{array}$   -----Eq(4)

$\text{constant} = >4B = 4$   -----Eq(5)

Solving Eq(2), Eq(3), Eq(4) and Eq(5) for $A, BC, D$ we get

$A = \text{1}/4, B = 1, C = -1/4andD = - \text{1}$

From Eq (1), the partial fraction decomposition of $\frac{x + 4}{x^2\left(x^2 + 4\right)}$ is $\frac{1}{4x} + \frac{1}{x^2} + \frac{\left( - \left(\frac{1}{4}\right)x - 1\right)}{x^2 + 4}$.