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If someone asked you to add 2x−1+3x+2\frac{2}{x-1} + \frac{3}{x+2}, you would combine them over a common denominator to get 5x+1(x−1)(x+2)\frac{5x+1}{(x-1)(x+2)}. But in calculus, you constantly need to do the exact opposite: split a messy rational fraction back into clean, simple building blocks.
This reverse process is called partial fraction decomposition, and a rational expression is simply a fraction where both the numerator and denominator are polynomials.
How do we systematically tear apart any rational fraction with distinct linear factors in the denominator?

Two Distinct Linear Factors

For every distinct linear factor in the denominator of a proper fraction (where the numerator's degree is strictly less than the denominator's degree), assign an unknown constant numerator: A,B,…A, B, \dots
Let's decompose 5x−1(x−3)(x+1)\frac{5x - 1}{(x - 3)(x + 1)} by writing the algebraic identity 5x−1(x−3)(x+1)≡Ax−3+Bx+1\frac{5x - 1}{(x - 3)(x + 1)} \equiv \frac{A}{x - 3} + \frac{B}{x + 1}.
Multiplying both sides by the common denominator (x−3)(x+1)(x - 3)(x + 1) clears all fractions completely, giving 5x−1=A(x+1)+B(x−3)5x - 1 = A(x + 1) + B(x - 3).
Under Method 1 (Strategic Substitution), substitute the real roots of each linear factor (such as setting x=−bax = -\frac{b}{a}) into the identity to eliminate all other terms and solve directly for the remaining unknown constant.
Under Method 2 (Equating Coefficients), expand the polynomials on the right-hand side and equate coefficients of corresponding powers of xx (x2x^2, x1x^1, x0x^0) to form and solve a system of simultaneous linear equations.
With Strategic Substitution, substitute the real root of each linear factor (such as setting x=−1x = -1 or x=3x = 3) into the identity to eliminate other terms and solve directly for each constant.
Alternatively, you can use Equating Coefficients: expand the right side to (A+B)x+(A−3B)(A + B)x + (A - 3B) and equate coefficients of corresponding powers of xx (x1x^1 and x0x^0) to solve the simultaneous system A+B=5A + B = 5 and A−3B=−1A - 3B = -1.
The Heaviside Cover-Up Rule provides an efficient shortcut for distinct linear factors: to find constant AA for factor (x−r)(x - r), evaluate P(r)Q′(r)\frac{P(r)}{Q'(r)} by covering up (x−r)(x - r) in the original fraction and substituting x=rx = r into the remainder.