Why Small Setup Errors Matter
A rational function can be difficult to integrate not because any one step is exceptionally complicated, but because an early setup error makes every later step unreliable. Skipping polynomial division, leaving out a repeated-factor term, or using an unsuitable numerator can all produce an expression that looks plausible while failing to equal the original function.
A reliable decomposition is an algebraic identity: after the terms are placed over a common denominator, the numerator must match the original numerator exactly. This gives a direct way to diagnose errors. Check the degree before decomposing, write the complete form required by the denominator, and recombine the result before using it in an integral.
Error 1: Skipping Polynomial Division
When the numerator's degree is at least the denominator's degree, partial fractions alone do not give the standard form. Polynomial division separates the rational function into a polynomial part and a proper rational remainder. The remainder can then be decomposed.
This division step is not a cosmetic rearrangement. A proper partial-fraction sum tends to zero as \(|x|\) grows, whereas an improper rational function may grow like a polynomial. A sum consisting only of proper partial-fraction terms therefore cannot represent an improper function with nonzero polynomial growth.
Worked Example: Divide Before Decomposing
Decompose \(\frac{x^3+1}{x^2-x}\). The numerator has degree \(3\), while the denominator has degree \(2\), so the fraction is improper.
Divide \(x^3+0x^2+0x+1\) by \(x^2-x\). The first quotient term is \(x\), since \(x^3/x^2=x\). Subtracting \(x(x^2-x)=x^3-x^2\) leaves \(x^2+0x+1\). The next quotient term is \(1\). Subtracting \(x^2-x\) leaves the remainder \(x+1\). Therefore
Now decompose the proper remainder. Write \(\frac{x+1}{x(x-1)}=\frac{A}{x}+\frac{B}{x-1}\). Multiplying by \(x(x-1)\) gives \(x+1=A(x-1)+Bx\). Setting \(x=0\) gives \(1=-A\), so \(A=-1\). Setting \(x=1\) gives \(2=B\), so \(B=2\). Thus
Check by recombining the fractional terms: \(-1/x+2/(x-1)=(-x+1+2x)/(x(x-1))=(x+1)/(x(x-1))\). Adding the quotient \(x+1\) recovers the division identity, so the result equals the original rational function.
A common wrong start is to try to write the entire original function as \(A/x+B/(x-1)\). Each term tends to zero as \(|x|\) grows, but the original function grows approximately like \(x\). No constants \(A\) and \(B\) can make that proposed form correct for all \(x\). The correction is to divide first and retain the polynomial quotient.
Error 2: Omitting Powers of a Repeated Factor
A repeated factor requires a term for each of its powers. If the denominator contains \((x-a)^m\), the partial-fraction form includes terms with denominators \((x-a),(x-a)^2,\ldots,(x-a)^m\). Leaving out one of them restricts the possible numerator after recombination, and can make an otherwise correct decomposition impossible.
Worked Example: Include Every Power
Decompose \(\frac{1}{(x-1)^2(x+2)}\). This is proper because the numerator has degree \(0\) and the denominator has degree \(3\). Since \(x-1\) is repeated, use the full form
Multiply by \((x-1)^2(x+2)\) to obtain the polynomial identity \(1=A(x-1)(x+2)+B(x+2)+C(x-1)^2\). Substitute \(x=1\): \(1=3B\), so \(B=1/3\). Substitute \(x=-2\): \(1=9C\), so \(C=1/9\). The coefficient of \(x^2\) on the right is \(A+C\); since the left has coefficient \(0\), \(A=-C=-1/9\). Hence
Check by recombining. The common numerator of the right-hand side, over \(9(x-1)^2(x+2)\), is \(-(x-1)(x+2)+3(x+2)+(x-1)^2\). Expanding gives \(-x^2-x+2+3x+6+x^2-2x+1=9\). The result is \(9/[9(x-1)^2(x+2)]\), exactly the original function.
If the \(\frac{B}{(x-1)^2}\) term is omitted, the proposed numerator after multiplying through is \(A(x-1)(x+2)+C(x-1)^2\). At \(x=1\), that numerator is \(0\), but the required numerator is \(1\). So the shorter form cannot work. The missing squared-denominator term is necessary, not optional.
Error 3: Using an Oversized Numerator
The numerator attached to a partial-fraction term must have degree strictly less than the degree of its denominator factor. Thus a linear factor \(x-a\) takes a constant numerator, and an irreducible quadratic takes a linear numerator. An oversized numerator may disguise polynomial growth or turn a concise coefficient problem into an unnecessarily large one.
Worked Example: Reduce an Oversized Quadratic Numerator
Consider \(\frac{x^2}{x^2+1}\). A tempting but unsuitable partial-fraction setup would treat this as a proper quadratic term over \(x^2+1\). It is not proper: both numerator and denominator have degree \(2\). In particular, a proposed proper-term numerator \(Ax+B\) cannot match \(x^2\), since the polynomial identity \(x^2=Ax+B\) would require the coefficient of \(x^2\) to be both \(1\) and \(0\).
Divide first. The leading terms give quotient \(1\), and subtracting \(x^2+1\) from \(x^2\) gives remainder \(-1\). Therefore
The remainder term is proper, and its numerator is constant, which is within the required linear bound. Check by recombining: \(1-1/(x^2+1)=(x^2+1-1)/(x^2+1)=x^2/(x^2+1)\).
It is possible to write \(x^2/(x^2+1)\) as a single fraction with a quadratic numerator; that equality is not itself false. The error is treating that oversized numerator as a proper partial-fraction term and proceeding as though the polynomial quotient had disappeared. Division exposes the polynomial part and leaves a numerator of the correct degree for the proper fraction.
Error 4: Losing a Term When Recombining
Finding coefficients is only half the job. A correct decomposition must still reproduce the original numerator when its terms are placed over a common denominator. A sign error, omitted term, or missed factor during this step can make a correct-looking answer wrong. Recombining is a direct check, not an optional extra.
Worked Example: Detect a Missing Repeated-Factor Term
For \(\frac{1}{x(x-1)^2}\), the complete form is \(\frac{A}{x}+\frac{B}{x-1}+\frac{C}{(x-1)^2}\). Multiplying by \(x(x-1)^2\) gives \(1=A(x-1)^2+Bx(x-1)+Cx\). Setting \(x=0\) gives \(A=1\); setting \(x=1\) gives \(C=1\). The coefficient of \(x^2\) on the right is \(A+B\), so \(B=-1\). Hence
Recombine all three terms. Their numerator over \(x(x-1)^2\) is \((x-1)^2-x(x-1)+x\). Expanding gives \(x^2-2x+1-x^2+x+x=1\), as required.
Now suppose the final term is accidentally dropped. The two remaining terms combine to \(\frac{1}{x}-\frac{1}{x-1}=\frac{x-1-x}{x(x-1)}=-1/[x(x-1)]\), which is not the original function. Even if the denominator is then adjusted to \(x(x-1)^2\), its numerator would be \(-(x-1)\), not \(1\). Recombining reveals the lost term immediately.
A Reliable Checking Routine
These errors can be caught with a few deliberate checks. Each one addresses a different point in the algebra:
If the numerator degree is at least the denominator degree, divide before decomposing.
For each repeated factor, include every power from the first up to the power in the denominator.
Use a constant over each linear factor and a linear polynomial over each irreducible quadratic factor.
Put the proposed terms over the original denominator and expand the numerator to verify the identity.
The last check is especially useful when an expression has many terms: it tests the whole decomposition, including signs and omitted powers, in one calculation. A decomposition that fails this identity must not be used to integrate the original function.
Why These Checks Matter for Integration
Partial fractions turn a complicated rational function into terms whose antiderivatives are familiar. But integration preserves the algebraic expression it is given; it cannot correct a faulty decomposition. A missing repeated-factor term, for example, changes the function before any integration begins. Likewise, an omitted polynomial quotient discards part of the integrand.
For that reason, verify the algebra first and integrate second. The essential test is equality of rational functions on their common domain, established by recombining and matching the numerator. This also clarifies the role of polynomial division: it preserves the original function exactly while separating the polynomial part from a proper rational remainder.
Check Your Understanding
For each question, identify the relevant check or write the required algebraic form.
- What degree comparison tells you that polynomial division is required?
- What terms must appear for a denominator containing \((x-3)^3\)?
- What is the maximum allowed numerator degree over an irreducible quadratic factor?
- Recombine \(\frac{1}{x}-\frac{1}{x-1}\). Does it equal \(\frac{1}{x(x-1)^2}\)?
- Why is a polynomial quotient lost if an improper rational function is represented only by proper partial-fraction terms?
