# Partial Fractions – Proper Fraction

### Partial Fractions – Proper Fraction

1. An expression of the form, f(x) = aₒ + a₁ x + a₂ x + … an xⁿ where n is a non-negative integer and aₒ, a₁, a₂, … an are real numbers such that an ≠ 0, is called a polynomial in x of degree n.

2. Division Algorithm: If f(x)/g(x), g(x) ≠ 0 are two polynomials, then \$ polynomial q(x), r(x) uniquely such that f(x) = g(x) q(x) + r(x) where r(x) = 0 or deg r(x) < deg g(x). The polynomial q(x) is called quotient and the polynomial r(x) is called remainder of f(x) when divided b yg(x).

3. Remainder Theorem: If f(x) is a polynomial, then the remainder of f(x) when divided by (x – a) is f(a).

4. Let f(x), g(x) be two polynomials. Then g(x) is said to divide f(x) or g(x) is said to be a divisor or factor of f(x) if there exists a polynomial q(x) such that f(x) = g(x) q(x).

5. Factor Theorem: If f(x) is a polynomial and f(a) = 0 then (x-a) is a factor of f(a).

6. If a rational function can be expressed as a sum of two ore more proper fractions, then each fraction is called a partial fraction of the given function.

7. Let f(x)/g(x) be a proper fraction.

If (ax + b)ⁿ where n ϵ N, is a factor of g(x) then the partial fractions corresponding to this factor are $$\frac{{{A}_{1}}}{ax+b}+\frac{{{A}_{2}}}{{{(ax+b)}^{2}}}+…..+\frac{{{A}_{n}}}{{{(ax+b)}^{n}}}$$, where A₁, A₂, …., An are constants.

If (ax² + bx + c )ⁿ where n ϵ N, is a factor of g(x), then the partial fractions corresponding to this factor are $$\frac{{{A}_{1}}x+{{B}_{1}}}{a{{x}^{2}}+bx+c}+\frac{{{A}_{2}}x+{{B}_{2}}}{{{(a{{x}^{2}}+bx+c)}^{2}}}+…..+\frac{{{A}_{n}}x+{{B}_{n}}}{{{(a{{x}^{2}}+bx+c)}^{n}}}$$, where A₁, A₂, …., An, B₁, B₂, …., Bn are constants.

Example 1: Resolve $$\frac{5x+2}{(1+3x)(1+2x)}$$ into partial fractions.

Solution: $$\frac{5x+2}{(1+3x)(1+2x)}$$

Let $$\frac{5x+2}{(1+3x)(1+2x)}=\frac{A}{1+3x}+\frac{B}{1+2x}$$ … (1)

Put x = -1/3 in equation (1)

$$A=\frac{5(-1/3)+2}{1+2(-1/3)}=\frac{1}{1}=1$$

A = 1

Put x = -1/2 in equation (1)

$$B=\frac{5(-1/2)+2}{1+3(-1/2)}=\frac{-1}{-1}=1$$

B = 1

$$\frac{5x+2}{(1+3x)(1+2x)}=\frac{1}{1+3x}+\frac{1}{1+2x}$$.

Example 2: Resolve $$\frac{1-x+6{{x}^{2}}}{x-{{x}^{3}}}$$ into partial fractions.

Solution: $$\frac{1-x+6{{x}^{2}}}{x-{{x}^{3}}}=\frac{1-x+6{{x}^{2}}}{x(1-{{x}^{2}})}$$

Let $$\frac{1-x+6{{x}^{2}}}{x(1-{{x}^{2}})}=\frac{1-x+6{{x}^{2}}}{x(1-x)(1+x)}$$

$$\frac{1-x+6{{x}^{2}}}{x(1-x)(1+x)}=\frac{A}{x}+\frac{B}{1-x}+\frac{C}{1+x}$$

$$\frac{1-x+6{{x}^{2}}}{x(1-x)(1+x)}=\frac{A(1-x)(1+x)+Bx(1+x)+Cx(1-x)}{x(1-x)(1+x)}$$

A (1 – x) (1 + x) + Bx(1 + x) + Cx(1 – x) = 1 – x + 6x² … (1)

Put x = 0 in equation (1)

A (1 – 0) (1 + 0 ) + 0 + 0 = 1 – 0 + 0

A = 1

Put x = 1 in equation (1)

0 + B (1) (1 + 1) + 0 = 1 – 1 + 6 (1)²

2B = 6

B = 3

Put x = -1 in equation (1)

0 + 0 + C (2) = 1 + 1 + 6(-1)²

-2C = 8

B = -4

∴ $$\frac{1-x+6{{x}^{2}}}{x-{{x}^{3}}}=\frac{1}{x}+\frac{3}{1-x}-\frac{4}{1+x}$$.