Chapter 11.5, Problem 36AYU

To determine

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

Answer to Problem 36AYU

Solution:

$\frac{1}{\left(x^2 + 16\right)} - \frac{15}{{\left(x^2 + 16\right)}^2}$

Explanation of Solution

Given:

$\frac{x^2 + 1}{{\left(x^2 + 16\right)}^2}$

Formula used:

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.

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

Calculation:

The given partial fraction decomposition takes the form

$\frac{x^2 + 1}{{\left(x^2 + 16\right)}^2} = \frac{Ax + B}{\left(x^2 + 16\right)} + \frac{Cx + D}{{\left(x^2 + 16\right)}^2}$

$= >x^2 + 1 = \left(Ax + B\right)\left(x^2 + 16\right) + Cx + D$   ----- Eq (1)

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

$x^3, = >A = 0$   -----Eq(2)

$x^2, = >B = 1$   -----Eq(3)

$\begin{array}{cc}x, = >& 16A + C = 0\end{array}$…   -----Eq(4)

$\begin{array}{cc}constant = >& 16B + D = 1\end{array}$…   -----Eq(5)

Solving for $A, B, C, D$ we get

$A = 0, B = 1, C = 0andD = \text{-} 15$

The partial fraction decomposition of $\frac{x^2 + 1}{{\left(x^2 + 16\right)}^2}$ is $\frac{1}{\left(x^2 + 16\right)} - \frac{15}{{\left(x^2 + 16\right)}^2}$.