Chapter 11.5, Problem 29AYU

In problems 13-46, find the partial fraction decomposition of each rational expression. $\frac{x^2+2x+3}{\left(x+1\right)\left(x^2+2x+4\right)}$

To determine

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

Answer to Problem 29AYU

Solution:

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

Explanation of Solution

Given:

$\frac{x2 + 2x + 3}{\left(x - 1\right)\left(x^2 + 2x + 4\right)}$

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

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

Where the numbers $A_1, A_2\mathrm{...}{A}_n$ are to be determined.

Calculation:

The given partial fraction decomposition takes the form

$\frac{x^2 + 2x + 3}{\left(x + 1\right)\left(x^2 + 2x + 4\right)} = \frac{A}{\left(x + 1\right)} + \frac{Bx + c}{\left(x^2 + 2x + 4\right)}$

$= >x^2 + 2x + 3 = A\left(x^2 + 2x + 4\right) + \left(Bx + C\right)\left(x + 1\right)$   -----Eq(1)

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

$\begin{array}{cc}{x}^2, & = >A + B = 1\end{array}$   -----Eq(2)

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

$\text{constant} = >4A + C + 3$   -----Eq(4)

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

$\text{A = 2/3,} \text{B = 1/3, C = 1/3}$

From Eq (1), the partial fraction decomposition of $\frac{x^2 + 2x + 3}{\left(x + 1\right)\left(x^2 + 2x + 4\right)}$ is $\frac{2}{3\left(x + 1\right)} + \frac{\left(\left(\frac{1}{3}\right)x + \frac{1}{3}\right)}{x^2 + 2x + 4}$.