Chapter 11.5, Problem 15AYU

To determine

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

Answer to Problem 15AYU

Solution:

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

Explanation of Solution

Given:

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

Formula used:

a. If $Q(x) = (x - a_1)\mathrm{...}\mathrm{...}(x - a_n)$ then partial fraction decomposition of $\text{P}(\text{x})/\text{Q}(\text{x})$ takes the form:

$\frac{P\left(x\right)}{Q\left(x\right)} = \frac{A_1}{\left(x - a_1\right)} + \frac{A_2}{\left(x - a_2\right)} + \dots \frac{A_n}{\left(x - a_n\right)}$

where $A_1, A_2\mathrm{...}{A}_n$ are to be determined.

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

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

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

Calculation:

It can be noticed that the denominator $x\left(x^2 + 1\right)$ has non repeated linear factor x and a non repeated irreducible quadratic factor $x^2 + 1$.

Therefore, the given partial fraction decomposition takes the form

$\begin{array}{rrr}\hfill \frac{1}{x\left(x^2 + 1\right)}& = \hfill & \frac{A}{x} + \frac{Bx + C}{\left(x^2 + 1\right)}\hfill \\ \hfill 1& \hfill = & A\left(x^2 + 1\right) + \left(Bx + C\right)x\hfill \\ \hfill & = \hfill & A{x}^2 + A + B{x}^2 + Cx\hfill \end{array}$…. Eq (1)

Equating the coefficients of $x^2$, $x$ and constant we get

$\begin{array}{r}\hfill A + B = 0\\ \hfill C = 0\\ \hfill A = 1\\ \hfill B = - 1\end{array}$

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