When we convert a fraction into a partial fraction, we are simplifying it so that it can be integrated or differentiated more easily.
[Case 1]
$$\frac{px + q}{(x-a)(x-b)} = \frac{A}{x-a} + \frac{B}{x-b}$$
[Case 2]
$$\frac{px + q}{(x-a)^2} = \frac{A}{x-a} + \frac{B}{(x-a)^2}$$
[Case 3]
$$\frac{px^2 + qx + r}{(x-a)(x^2 + bx + c)} = \frac{A}{x-a} + \frac{Bx + C}{x^2 + bx + c}$$
As you can see, there are three possible cases for partial fractions. Let's work through examples of each one-by-one.
[Case 1]
$$\frac{11x + 25}{(x+2)(x+3)} = \frac{A}{x+2} + \frac{B}{x+3} = \frac{A(x+3) + B(x+2)}{(x+2)(x+3)}$$
$$A(x+3) + B(x+2) = 11x + 25$$
$$Ax + 3A + Bx + 2B = 11x + 25$$
$$(A+B)x + (3A+2B) = 11x + 25$$
Equation 1 -> \( A + B = 11 \)
Equation 2 -> \( 3A + 2B = 25 \)
Simultaneous linear equations gives us
\( A = 3 \), \( B = 8 \)
[Case 2]
$$\frac{x+3}{(x-3)^2} = \frac{A}{(x-3)} + \frac{B}{(x-3)^2} = \frac{A(x-3) + B}{(x-3)^2}$$
$$A(x-3) + B = x+3$$
$$Ax - 3A + B = x+3$$
$$Ax + (-3A+B) = x+3$$
Equation 1: \( A = 1 \)
Equation 2: \( -3A + B = 3 \)
Solving simultaneous linear equations gives us
\( A = 1 \), \( B = 6 \)
[Case 3]
$$\frac{5x^2 - x + 4}{(x-3)(x^2+3x+5)} = \frac{A}{(x-3)} + \frac{Bx+C}{(x^2+3x+5)} = \frac{A(x^2+3x+5) + (x-3)(Bx+C)}{(x-3)(x^2+3x+5)}$$
$$Ax^2 + 3Ax + 5A + Bx^2 + Cx - 3Bx - 3C = 5x^2 - x + 4$$
$$(A+B)x^2 + (3A-3B+C)x + (5A-3C) = 5x^2 - x + 4$$
Equation 1: \( A+B = 5 \)
Equation 2: \( 3A-3B+C = -1 \)
Equation 3: \( 5A-3C = 4 \)
Solving simultaneous linear equations gives us
\( A = 2 \), \( B = 3 \), \( C = 2 \)
Important note: The concept remains the same even for situations like \(\frac{px+q}{(x-a)(x-b)(x-c)}\) or \(\frac{px-q}{(x-a)(x-b)(x-c)^2}\). Just modify the partial fraction accordingly to
\(\frac{A}{(x-a)} + \frac{B}{(x-b)} + \frac{C}{(x-c)}\) and \(\frac{A}{(x-a)} + \frac{B}{(x-b)} + \frac{C}{(x-c)} + \frac{D}{(x-c)^2}\) respectively.
[Case 4: Improper Fractions]
Now, what happens when an improper fraction is used?
If the highest power of the denominator is less than or equal to the highest power of the numerator, the fraction is an improper fraction.
Example:
$$\frac{5x^2 - x + 4}{(x-3)(x+5)} = \frac{5x^2 - x + 4}{x^2 + 2x - 15}$$
Highest power in numerator is 2 from \( 5x^2 \). Highest power in denominator is 2 from \( x^2 \). As the highest power in the numerator is not less than the denominator, this is an improper fraction.
Here's how we solve it:
(1) Divide the numerator by the denominator.
$$\begin{array}{r} x^2 + 2x - 15 \overline{)5x^2 - x + 4} \end{array} \\ \underline{-(5x^2 + 10x - 75)} \\ -11x + 79$$
$$5 + \frac{-11x + 79}{x^2 + 2x - 15} = 5 + \frac{-11x + 79}{(x-3)(x+5)}$$
(2) Split \( \frac{-11x + 79}{(x-3)(x+5)} \) into partial fractions.
$$\frac{A}{x-3} + \frac{B}{x+5} = \frac{-11x + 79}{(x-3)(x+5)}$$
A(x+5) + B(x-3) = -11x + 79
Ax + 5A + Bx - 3B = -11x + 79
(A+B)x + (5A - 3B) = -11x + 79
A + B = -11
5A - 3B = 79
Solving simultaneous equations, we get
$$A = \frac{23}{4}, B = -\frac{67}{4}$$
Thus, full partial fraction is \( 5 + \frac{23/4}{x-3} + \frac{-67/4}{x+5} \)