Partial Fractions Are Not That Hard, Actually
The hardest part about Decomposing Into Partial Fractions is the factoring step. Once your denominator is fully factored, the decomposition itself is basically a pattern-matching exercise with a single trick to learn. I spent far too many afternoons staring at denominators that refused to factor, wondering why the rest of the problem seemed impossible. Take your rational function. Make sure the numerator degree is strictly less than the denominator degree. If it is not, do polynomial long division first. This is where most students lose points because they skip straight to the decomposition and end up with garbage coefficients. After long division, you separate the polynomial quotient (which is easy to integrate or transform on its own) from the proper rational remainder. Then you factor the denominator completely. Not over the reals. Over the rationals. This means linear factors and irreducible quadratic factors with integer coefficients. If your denominator is $x^3 - 1$, that factors as $(x-1)(x^2 + x + 1)$, not as $(x-1)(x - e^{2\pi i/3})(x - e^{-2\pi i/3})$. You are working in real coefficients. Every time someone tries to decompose with complex roots in a calculus class, the answer gets wrong or unnecessarily complicated.
Once the denominator is factored, you write the decomposition form. Distinct linear factors each get a constant numerator. Repeated linear factors each get their own term. An irreducible quadratic factor gets a linear numerator $Bx + C$. A repeated irreducible quadratic gets terms for each power. The number of unknowns equals the total count of factors when you account for multiplicity. Here is an example. Take $\frac{3x^2 + 2x - 1}{(x-1)(x+2)}$. The numerator and denominator are both degree 2, so first I divide: $3x^2 + 2x - 1$ divided by $x^2 + x - 2$ gives quotient $3$ and remainder $-x + 5$. So the expression becomes $3 + \frac{-x + 5}{(x-1)(x+2)}$. Now the proper fraction: $\frac{-x + 5}{(x-1)(x+2)} = \frac{A}{x-1} + \frac{B}{x+2}$. Multiply through by $(x-1)(x+2)$ and you get $-x + 5 = A(x+2) + B(x-1)$. Set $x = 1$: $4 = 3A$, so $A = 4/3$. Set $x = -2$: $7 = -3B$, so $B = -7/3$. The full decomposition is $3 + \frac{4/3}{x-1} - \frac{7/3}{x+2}$. Done. That cover-up trick works reliably for distinct linear factors. Set the variable equal to the root of the factor you want, evaluate the remaining expression, and solve. It cuts the algebra in half. For repeated factors, the cover-up method still gives you the coefficient for the highest power, but you still need to solve for the rest.
Consider $\frac{2x + 3}{(x-1)^2(x+2)}$. Write it as $\frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{x+2}$. Multiply through: $2x + 3 = A(x-1)(x+2) + B(x+2) + C(x-1)^2$. Set $x = 1$: $5 = 3B$, so $B = 5/3$. Set $x = -2$: $-1 = 9C$, so $C = -1/9$. For $A$, pick any convenient value like $x = 0$: $3 = A(-1)(2) + \frac{5}{3}(2) + (-\frac{1}{9})(1)$. Solve and get $A = 1/9$. Verify by combining all terms back into a single fraction. If the numerator matches your original, you did not make an arithmetic mistake.
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Repeated Irreducible Quadratics
This is where people routinely mess up. Take $\frac{x}{(x^2+1)^2}$. The decomposition form is $\frac{Ax+B}{x^2+1} + \frac{Cx+D}{(x^2+1)^2}$. Two terms, four unknowns. Multiply through and equate coefficients. Set $x = 0$ to get one equation quickly. Then compare the $x^3$ coefficient, the $x^2$ coefficient, and the constant term. You get $A = 0$, $B = 0$, $C = 1$, $D = 0$. The decomposition is just $\frac{x}{(x^2+1)^2}$. Which means the original expression was already in the right form, but this shows the process does not always save you work. Sometimes it confirms you already had the answer. I spent two weeks last year debugging a circuit analysis problem where my partial fraction decomposition kept producing slightly wrong residues. The transfer function had a repeated quadratic factor with a near-degenerate root pair. When I solved for the coefficients by hand, I was off by about 0.03 in the imaginary part. The numerical simulation disagreed with the analytical result. The fix was not a clever trick. I rewrote the problem as a system of linear equations in the unknown coefficients, solved it using row reduction instead of substitution, and caught the arithmetic error immediately. Substitution works fine for textbook problems with clean numbers. Real problems do not always cooperate. Another thing that trips people up: improper fractions with the same degree numerator and denominator. Some students treat this as a decomposition problem and never do the long division. The result is mathematically valid but completely useless for integration or inverse Laplace transforms because the "constant" term buried in the coefficients messes up every subsequent step. Do the division first. Always.
There are cases where partial fraction decomposition simply will not help you. If your denominator is a high-degree polynomial with no rational roots, like $x^5 + x + 1$, factorization over the rationals may be impossible or require finding roots numerically. In that scenario, you either accept a numerical approximation for the poles and proceed with a numerical residue calculation, or you leave the expression as-is and use a different method entirely. For integral evaluation, numerical quadrature is faster. For Laplace inversion, a table lookup or a computational tool is more practical. Partial fractions assumes you can factor the denominator. When you cannot, the method stops being useful.
Common Pitfalls
Forgetting to check whether the rational function is proper before starting. Setting the decomposition form incorrectly for repeated factors by omitting lower powers. Using complex roots when real coefficients are required. Making sign errors when clearing denominators. Skipping the verification step and assuming the coefficients are correct because the algebra looked clean. The verification step takes about 30 seconds and prevents hours of confusion later. Also worth noting: the method only works for rational functions. If your expression involves logarithms, exponentials, or trigonometric terms in the denominator, partial fraction decomposition does not apply directly. You may need a substitution first to convert the expression into a rational function of a new variable, but that is a separate problem with its own set of constraints. The decomposition process itself is mechanical once you know the forms. The real skill is recognizing what form applies to what denominator structure and catching mistakes before they compound through the rest of the problem. Factor carefully. Write the correct form. Use the cover-up method where it applies. Fall back to coefficient comparison when it does not. Verify. Move on.
