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