Why Degree Comparison Comes First
A rational integrand is a quotient of polynomials. Its denominator may have factors that suggest familiar antiderivative forms, but those factors do not by themselves tell you whether partial fractions can represent the whole quotient. First compare the degrees of the numerator and denominator.
If the numerator has degree at least as large as the denominator, the rational function is improper. In that case, polynomial division separates it into a polynomial and a rational function with a lower-degree numerator. That separation matters because the usual partial-fraction decomposition applies to the proper remainder, not to the original improper expression alone.
For a nonzero polynomial, its degree is the exponent of its highest-power term with a nonzero coefficient. For example, \(4x^3-x+2\) has degree \(3\). When comparing degrees, look at the polynomials after any valid simplification or cancellation.
The Division Step
Polynomial division gives polynomials \(Q(x)\) and \(R(x)\) such that
where either \(R(x)=0\) or \(\deg R<\deg D\). Dividing by \(D(x)\), wherever \(D(x)\ne 0\), gives
If \(P/D\) is improper, this is the useful rewrite: \(Q\) is a polynomial, and \(R/D\) is proper. The polynomial part can be integrated using the power rule. If the proper remainder has a denominator that factors suitably, partial fractions can then be applied to that remainder.
Why the Method Works
The key is the division identity, not a change to the value of the integrand. On the domain where \(D(x)\ne 0\), the identity \(P=DQ+R\) guarantees that the original quotient equals the sum of the quotient polynomial and the remainder fraction. Linearity of integration then lets us handle these parts separately.
Determine whether the numerator degree is at least the denominator degree.
Write the numerator as denominator times quotient, plus a remainder of lower degree.
Apply the power rule term by term.
When its denominator factors into distinct linear factors, decompose it into constant multiples of reciprocals of those factors, then integrate.
Differentiate the proposed antiderivative and compare it with the original integrand on its domain.
For instance, if the proper remainder has denominator \((x-a)(x-b)\), with \(a\ne b\), a decomposition of the form \(A/(x-a)+B/(x-b)\) can be tested by multiplying through by the denominator and matching coefficients. This form is appropriate for a proper remainder; if the original numerator degree is too high, the quotient polynomial must not be omitted.
Worked Examples
Worked Example: Divide Before Integrating a Linear Denominator
Find \(\int \frac{x^2+3x+5}{x+1}\,dx\).
The numerator has degree \(2\), while the denominator has degree \(1\). The rational function is improper, so divide:
Therefore, for \(x\ne -1\),
Integrate each term:
To check, differentiate the proposed antiderivative:
The derivative matches the original integrand wherever \(x\ne -1\).
Worked Example: Divide, Then Use Partial Fractions
Find \(\int \frac{2x^3+3x^2}{x^2-1}\,dx\).
The numerator degree is \(3\), greater than the denominator degree \(2\). Polynomial division gives
Thus the quotient and remainder are both \(2x+3\), and
The remainder fraction is proper. Since \(x^2-1=(x-1)(x+1)\), write
Multiplying by \((x-1)(x+1)\) gives \(2x+3=A(x+1)+B(x-1)\). Matching coefficients yields \(A+B=2\) and \(A-B=3\). Adding these equations gives \(2A=5\), so \(A=5/2\); then \(B=2-5/2=-1/2\). Therefore,
Integrating term by term gives
Differentiate to check:
The check holds on the domain \(x\ne -1,1\).
Worked Example: Equal Degrees Still Require Division
Find \(\int \frac{x^2+2x+3}{x^2+x}\,dx\).
Both numerator and denominator have degree \(2\). Because the degrees are equal, the rational function is improper. Divide:
Consequently,
For the proper remainder, set \(\frac{x+3}{x(x+1)}=\frac{A}{x}+\frac{B}{x+1}\). Clearing the denominator gives \(x+3=A(x+1)+Bx\). Comparing constant terms gives \(A=3\), and comparing coefficients of \(x\) gives \(A+B=1\), so \(B=-2\). Hence
Integrating produces
Differentiate and combine the terms over a common denominator:
This verifies the result for \(x\ne 0,-1\).
A Common Mistake: Trying Partial Fractions Too Soon
A frequent error is to try to represent an improper rational function using only the reciprocal factors of its denominator. For the first example, one might incorrectly try to write
Multiplying through by \(x+1\) would require \(x^2+3x+5=A\), which cannot hold for every \(x\): the left side is a quadratic polynomial and the right side is constant. The error is not a poor choice of \(A\); the proposed form is missing a polynomial part.
Polynomial division supplies precisely that missing part:
Only the proper remainder is represented by the reciprocal factor. Keep the quotient polynomial as part of the decomposition and integrate it as well.
When Division Is Not Needed
If the numerator degree is already less than the denominator degree, the rational function is proper. For example, \((2x+1)/(x^2-1)\) has numerator degree \(1\) and denominator degree \(2\). It is already in the right degree form for partial fractions, so polynomial division would add no useful step.
The test concerns degree, not the number of terms in either polynomial. A numerator with many terms can still have lower degree than the denominator; conversely, a short numerator can have degree at least as large. Also, a denominator factorization does not remove the need to check degrees first.
Why This Strategy Matters
Degree comparison is a small preliminary step that prevents a structural error. When division is required, it exposes a polynomial part that would otherwise be concealed in the rational quotient. It also leaves a proper remainder, making the subsequent decomposition easier to organize. The resulting pieces often have standard antiderivatives: powers for the polynomial terms and logarithms for the simple reciprocal factors.
Check Your Understanding
For each question, decide what the degree comparison implies before attempting any integration.
- Is \((3x^2+1)/(x^2+4)\) proper or improper? Should polynomial division come first?
- What degree condition must the remainder satisfy after dividing \(P(x)\) by \(D(x)\)?
- Divide \(x^2+4x+7\) by \(x+2\), and write the resulting quotient-plus-remainder expression for the rational function.
- Why can a decomposition using only constants over linear denominator factors fail for an improper rational function?
- After dividing an improper rational function, what should you check before applying partial fractions to the remainder?
