Chapter 11.5, Problem 16AYU

In problems 13-46, find the partial fraction decomposition of each rational expression. $\frac{1}{\left(x+1\right)\left(x^2+4\right)}$

To determine

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

Answer to Problem 16AYU

Solution:

$\frac{1}{5\left(x + 1\right)} + \frac{\left( - \frac{1}{5}\right)x + \frac{1}{5}}{\left(x^2 + 4\right)}$.

Explanation of Solution

Given:

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

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 A₁, A₂…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 $\left(x + 1\right)\left(x^2 + 4\right)$ has non repeated linear factor $(x+1)$ and a non repeated irreducible quadratic factor $x^2 + 4$.

Therefore, the given partial fraction decomposition takes the form

$\begin{array}{rrr}\hfill \frac{1}{\left(x + 1\right)\left(x^2 + 4\right)}& = \hfill & \frac{A}{x + 1} + \frac{Bx + C}{\left(x^2 + 4\right)}\hfill \\ \hfill 1& \hfill = & A\left(x^2 + 4\right) + \left(Bx + C\right)\left(x + 1\right)\hfill \\ \hfill & = \hfill & A{x}^2 + 4A + B{x}^2 + Bx + Cx + C\hfill \end{array}$…. Eq (1)

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

$\begin{array}{r}\hfill A + B = 0\\ \hfill B + C = 0\\ \hfill 4A + C = 1\end{array}$

Solving for A,B and C we get

$\begin{array}{r}\hfill C = 1/5\\ \hfill A = 1/5\\ \hfill B = - 1/5\end{array}$

From Eq (1), the partial fraction decomposition of $\frac{1}{\left(x + 1\right)\left(x^2 + 4\right)}$ is $\frac{1}{5\left(x + 1\right)} + \frac{\left( - \frac{1}{5}\right)x + \frac{1}{5}}{\left(x^2 + 4\right)}$.