Chapter 11.5, Problem 28AYU

To determine

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

Answer to Problem 28AYU

Solution:

$\frac{14}{3\left(x-1\right)}+\frac{4}{{\left(x-1\right)}^2}+\frac{\left(-\left(\frac{14}{3}\right)x+4/3\right)}{x^2+2}$

Explanation of Solution

Given:

$\frac{10x^2+2x}{{\left(x-1\right)}^2\left(x^2+2\right)}$

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 $A$ and $B$ are to be determined.

b. If $Q$ has repeated linear factor of the form ${\left(x-a\right)}^n$, $n\ge 2$ an integer, then, in the partial fraction decomposition of $P\left(x\right)/Q\left(x\right)$, takes the form:

$\frac{P\left(x\right)}{Q\left(x\right)}=\frac{A_1}{\left(x-a\right)}+\frac{A_2}{{\left(x-a\right)}^2}+\dots +\frac{A_n}{{\left(x-a\right)}^n}$

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

Calculation:

The given partial fraction decomposition takes the form

$\frac{10x^2+2x}{{\left(x-1\right)}^2\left(x^2+2\right)}=\frac{A}{\left(x-1\right)}+\frac{B}{{\left(x-1\right)}^2}+\frac{Cx+D}{\left(x^2+2\right)}$

$\Rightarrow 10x^2+2x=A\left(x-1\right)\left(x^2+2\right)+B\left(x^2+2\right)+\left(Cx+D\right){(}^x$    ----- Eq (1)

$=A\left(x^3-2-x^2+2x\right)+B\left(x^2+2\right)+\left(Cx+D\right)\left(x^2-2x+1\right)$

$=\left(x^3-x^2+2x-2\right)+B\left(x^2+2\right)+C\left(x^3-2x^2+x\right)+D\left(x^2-2x+1\right)$

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

$x{,}^3=>A+C=0$    ----- Eq (2)

$x^2,=>-A+B-2C+D=10$    ----- Eq (3)

$x,=>2A+C-2D=2$    ----- Eq (4)

constant $=>-2A+2B+D=0$    ----- Eq (5)

$A=14/3,B=4,C=-14/3\text{ and }D=4/3$

From Eq (1), the partial fraction decomposition of $\frac{x+4}{x^2\left(x^2+4\right)}$ $\frac{14}{3\left(x-1\right)}+\frac{4}{{\left(x-1\right)}^2}+\frac{\left(-\left(\frac{14}{3}\right)x+4/3\right)}{x^2+2}$.