Why Divide Before Integrating?
An improper rational function is a quotient of polynomials in which the numerator's degree is at least the denominator's degree. Partial fractions are designed for a proper rational function, where the numerator's degree is smaller. Polynomial division resolves the mismatch: it separates the integrand into a polynomial and a proper rational remainder.
That separation is useful for integration because the polynomial part can be integrated directly, one power at a time. The proper remainder can then be expressed using partial fractions, and its terms can be integrated using the appropriate antiderivative formulas. The final answer combines both parts.
The Quotient-Remainder Identity
Let \(P(x)\) and \(Q(x)\) be polynomials, with \(Q\) not the zero polynomial. Polynomial division gives unique polynomials \(S(x)\) and \(R(x)\) such that
Where \(Q(x)\ne 0\), dividing this identity by \(Q(x)\) yields
The polynomial \(S(x)\) is the quotient, and \(R(x)/Q(x)\) is the proper rational remainder when \(R\ne 0\). If the remainder is then decomposed into partial fractions, the original integral becomes the sum of the integral of \(S(x)\) and the integrals of those partial-fraction terms.
Why the Terms Can Be Integrated Separately
The key fact is the linearity of differentiation. If \(F'(x)=f(x)\) and \(G'(x)=g(x)\), then \((F+G)'=f+g\). Multiplying an antiderivative by a constant works in the same way: \((cF)'=cf\). Therefore, once an integrand has been expressed as a sum of terms, we can find an antiderivative of each term and add the results.
For the polynomial part, the power rule supplies the needed antiderivatives. For example, when \(k\) is a nonnegative integer, the derivative of \(x^{k+1}/(k+1)\) is \(x^k\). Thus each monomial in the quotient has a direct antiderivative. The partial-fraction terms are handled separately using their standard antiderivatives; the examples below use simple linear factors.
Worked Examples
Worked Example: A Linear Quotient and One Partial Fraction
Integrate \(\frac{x^2+3x+5}{x+1}\). The numerator has degree \(2\), while the denominator has degree \(1\), so first divide. The leading quotient term is \(x\), since \(x^2/x=x\). Multiplying gives \(x(x+1)=x^2+x\). Subtracting from the numerator leaves \(2x+5\). The next quotient term is \(2\), since \(2x/x=2\). Multiplying gives \(2(x+1)=2x+2\), leaving remainder \(3\). Therefore,
Integrate each term. The polynomial terms give \(\int x\,dx=x^2/2\) and \(\int 2\,dx=2x\). Since the derivative of \(\ln|x+1|\) is \(1/(x+1)\) wherever \(x\ne-1\), the remaining term gives \(3\ln|x+1|\). Thus
Check by differentiation: the derivative of the proposed answer is \(x+2+3/(x+1)\). Recombining over the common denominator gives \[ x+2+\frac{3}{x+1} =\frac{(x+2)(x+1)+3}{x+1} =\frac{x^2+3x+2+3}{x+1} =\frac{x^2+3x+5}{x+1}. \] This is the original integrand. The result holds on either interval that does not contain \(x=-1\), with the constant chosen on that interval.
Worked Example: A Quotient and Two Distinct Linear Factors
Integrate \(\frac{x^3+2x^2+5x+4}{x^2+x-2}\). Factor the denominator as \(x^2+x-2=(x-1)(x+2)\). For the division, the first quotient term is \(x\), because \(x^3/x^2=x\). Subtracting \(x(x^2+x-2)=x^3+x^2-2x\) leaves \(x^2+7x+4\). The next quotient term is \(1\). Subtracting \(x^2+x-2\) leaves \(6x+6\). Hence
The proper remainder is \(\frac{6x+6}{(x-1)(x+2)}\). Write it as \(A/(x-1)+B/(x+2)\). Clearing the denominators gives \(6x+6=A(x+2)+B(x-1)\). Setting \(x=1\) gives \(12=3A\), so \(A=4\). Setting \(x=-2\) gives \(-6=-3B\), so \(B=2\). Therefore,
Integrate the polynomial quotient term by term: \(\int x\,dx=x^2/2\) and \(\int 1\,dx=x\). The two simple linear terms give logarithms, so
For a check, differentiate: the result is \(x+1+4/(x-1)+2/(x+2)\). Combining the fractions, \[ \frac{4}{x-1}+\frac{2}{x+2} =\frac{4(x+2)+2(x-1)}{(x-1)(x+2)} =\frac{6x+6}{x^2+x-2}. \] Adding \(x+1\) over the same denominator gives numerator \((x+1)(x^2+x-2)+6x+6=x^3+2x^2+5x+4\), as required. The original denominator vanishes at \(1\) and \(-2\), so the antiderivative is considered on intervals that do not cross either value.
Worked Example: A Quadratic Quotient with a Proper Remainder
Integrate \(\frac{x^4-2x^3+2x^2+5x+4}{x^2-1}\). Divide the leading terms: \(x^4/x^2=x^2\). Subtracting \(x^2(x^2-1)=x^4-x^2\) leaves \(-2x^3+3x^2+5x+4\). The next quotient term is \(-2x\). Subtracting \(-2x(x^2-1)=-2x^3+2x\) leaves \(3x^2+3x+4\). The next quotient term is \(3\). Subtracting \(3(x^2-1)=3x^2-3\) leaves \(3x+7\). Thus
Since \(x^2-1=(x-1)(x+1)\), decompose the proper remainder as \(A/(x-1)+B/(x+1)\). Clearing denominators gives \(3x+7=A(x+1)+B(x-1)\). Setting \(x=1\) gives \(10=2A\), so \(A=5\). Setting \(x=-1\) gives \(4=-2B\), so \(B=-2\). Therefore the integrand is \(x^2-2x+3+5/(x-1)-2/(x+1)\).
Integrating the quotient term by term gives \(x^3/3-x^2+3x\). Integrating the two simple linear terms gives the combined antiderivative
Differentiate to check: the derivative is \(x^2-2x+3+5/(x-1)-2/(x+1)\). The fractional part recombines as \[ \frac{5(x+1)-2(x-1)}{x^2-1} =\frac{3x+7}{x^2-1}. \] Combining this with the quotient reproduces the division identity above, and hence the original integrand. The denominator excludes \(x=-1\) and \(x=1\).
Constants and the Domain of an Antiderivative
An indefinite integral describes a family of antiderivatives. If two antiderivatives of the same function are defined on a connected interval, they differ by a constant. For rational functions, denominator zeros are excluded from the domain. These excluded values may divide the domain into separate intervals, and the constant can be chosen independently on each interval.
For example, an expression involving \(\ln|x-1|\) is defined for \(x<1\) and for \(x>1\), but not at \(x=1\). An antiderivative formula may be written on either interval; it does not make the original rational function defined at the excluded point. Keeping track of these restrictions is part of stating the result correctly.
A Common Error: Dropping the Quotient
In the first worked example, the division gives \(x+2+3/(x+1)\). If the quotient is mistakenly discarded, one might write \(3\ln|x+1|+C\) as the answer. Differentiating that expression gives only \(3/(x+1)\), not the original integrand. The correction is to integrate every part of the division identity: \(x\) contributes \(x^2/2\), \(2\) contributes \(2x\), and the remainder contributes \(3\ln|x+1|\).
A related error is to integrate \(x\) as \(\ln|x|\). That is incorrect because \(\frac{d}{dx}\ln|x|=1/x\), while \(\frac{d}{dx}(x^2/2)=x\). Differentiating the proposed antiderivative is an efficient way to catch either mistake.
A Reliable Sequence of Steps
Find \(S\) and \(R\) with \(P=QS+R\) and \(R=0\) or \(\deg R<\deg Q\).
Use partial fractions on \(R/Q\) when the remainder is nonzero.
Apply the power rule to every quotient term and integrate the partial-fraction terms.
Add \(C\) and exclude any values that make the original denominator zero.
Differentiate the complete answer and check that it equals the original rational function.
Check Your Understanding
Use the quotient-remainder identity to organize each integral. When checking, differentiate the complete antiderivative and compare it with the original integrand.
- What condition on the degrees of \(R\) and \(Q\) makes \(R/Q\) a proper rational function?
- For a quotient \(S(x)=3x^2-4x+5\), what is its term-by-term antiderivative?
- Why does omitting a nonzero polynomial quotient generally produce an incorrect antiderivative?
- Where is an antiderivative involving \(\ln|x-2|\) excluded from its domain?
- What does differentiating a proposed answer verify in an integration problem?
