Inequalities

The process of decomposition of fraction depends on one of the following:

1. To every linear factor $ax + b$ in the denominator there corresponds a fraction of the form

$\frac{A}{ax + b}$

2. To every repeated factor like $(ax + b)^n$ in the denominator there corresponds $n$ fractions of the form

$\frac{A_1}{ax + b} + \frac{A_2}{(ax + b)^2} + \cdots + \frac{A_n}{(ax + b)^n}$

3. To every factor of the form $a_n x^n + a_{n-1} x^{n-1} + \ldots + a_1 x + a_0$ in the denominator there correspond fraction of the form

$\frac{A_1 x + B_1}{a_1 x^2 + b_1 x + c_1} + \frac{A_2 x + B_2}{a_2 x^2 + b_2 x + c_2} + \cdots$

4. If the degree of the numerator is greater than or equal to the degree of denominator, division is encouraged and the remainder is treated as in (1), (2) or (3)

Examples

1. Express in partial fraction

$\frac{3x+7}{x^2+2x-8}$

Solution

Let $\frac{3x+7}{x^2+2x-8} = \frac{3x+7}{(x-2)(x+4)} = \frac{A}{x-2} + \frac{B}{x+4}$

Where A and B are constant

$3x + 7 = A(x+4) + B(x-2)$

$3x + 7 = Ax + 4A + Bx - 2B$

$3x + 7 = (A+B)x + 4A - 2B$

Equating coefficients:

$3 = A + B \ldots (i)$

$7 = 4A - 2B \ldots (ii)$

2(i) + (ii) gives

$13 = 6A \implies A = \frac{13}{6}$

From (i)

$B = \frac{5}{6}$

$\frac{3x+7}{x^2+2x-8} = \frac{13}{6(x-2)} + \frac{5}{6(x+4)}$

2. $\frac{x^2+1}{(x-1)(x+1)^3}$

Solution

Let $\frac{x^2+1}{(x-1)(x+1)^3} = \frac{A}{x-1} + \frac{B}{x+1} + \frac{C}{(x+1)^2} + \frac{D}{(x+1)^3}$

$x^2 + 1 \equiv A(x + 1)^3 + B(x + 1)^2 (x - 1) + C(x + 1)(x - 1) + D(x - 1)$

When $x = 1$

$2 = 8A \rightarrow A = \frac{1}{4}$

When $x = -1$

$2 = -2D \rightarrow D = -1$

When $x = 0$

$1 = A - B - C - D$

$1 = \frac{1}{4} - B - C - (-1)$

$1 = 4B + 4C \ldots (i)$

When $x = -2$

$5 = -A - 3B + 3C - 3D$

$5 = -\frac{1}{4} - 3B + 3C + 3$

$9 = -12B + 12C$