Chapter 11.5, Problem 24AYU

To determine

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

Answer to Problem 24AYU

Solution:

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

Explanation of Solution

Given:

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

Formula used:

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 $P\left(x\right)\text{/}Q\left(x\right)$, 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 denominator contains the repeated linear factors ${\left(x\right)}^2 \& {\left(x - 2\right)}^2$

The given partial fraction decomposition takes the form

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

$\Rightarrow \text{x} + 1 = Ax{\left(x - 2\right)}^2 + B{\left(x - 2\right)}^2 + C{x}^2\left(x - 2\right) + D{x}^2$   -----Eq (1)

Let $x = 2$ in Eq(1)

$\text{D} = 3\text{/}4$

Let $x = 0$ in Eq(1)

$\text{B} = 1/4$

Expanding Eq (1)

$\begin{array}{lll}x + 1\hfill & = \hfill & Ax\left(x^2 + 4 - 4x\right) + B\left(x^2 + 4 - 4x\right) + C{x}^3 - 2C{x}^2 + D{x}^2\hfill \\ \hfill & = \hfill & A\left(x^3 - 4x^2 + 4x\right) + B\left(x^2 - 4x + 4\right) + C\left(x^3 - 2x^2\right) + D{x}^2\hfill \end{array}$   -----Eq(2)

Equating the co-efficients on both sides in Eq(2),

$x^3, \Rightarrow A + C = 0$   -----Eq(3)

$x^2, \Rightarrow - = - 4A + B - 2C + D = 0$   -----Eq(4)

$x, \Rightarrow 4A - 4B = 1$   -----Eq(5)

$constant\Rightarrow 4B = 1$   -----Eq(6)

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

$A = 1/2, B = 1/4, C = - 1/2 and D = 3/4$

From Eq (1), the partial fraction decomposition of

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