Why Decompose a Rational Function?
A rational function is a quotient of polynomials. When its denominator is a product of distinct linear factors, a single fraction may be difficult to integrate directly. Partial fractions rewrite it as a sum of simpler fractions, each of which has a logarithmic antiderivative.
The method applies when the rational function is proper: the degree of the numerator is less than the degree of the denominator. If it is not proper, first use polynomial division so that the remaining rational part is proper. The denominator must also be factored into distinct linear factors over the real numbers for the method in this tutorial.
There is one constant numerator for each distinct linear factor. Once the constants are found, each term can be integrated using the formula for the logarithmic antiderivative of \(1/(x-a)\). The original function is undefined at the values \(a_1,\dots,a_n\), so an antiderivative is understood on an interval that does not contain any of those values.
Finding the Coefficients
To find the constants, first write the proposed decomposition. Then multiply both sides by the original denominator. This clears the fractions and gives a polynomial identity. Substituting a denominator zero often makes all but one term vanish, giving a direct equation for one coefficient.
Confirm that the fraction is proper and that its denominator is a product of distinct real linear factors.
Use an unknown constant numerator above each distinct linear factor.
Multiply through by the full denominator to obtain a polynomial identity.
Substitute convenient values, or compare coefficients, then recombine the fractions to verify the result.
Use logarithms with absolute values, and restrict the antiderivative to an interval containing no denominator zero.
Worked Example: Decompose and Integrate Two Factors
Decompose and integrate \(\frac{3x+5}{(x+1)(x+2)}\).
The numerator has degree \(1\), and the denominator has degree \(2\), so the fraction is proper. Its two distinct factors suggest the form
Multiply both sides by \((x+1)(x+2)\):
Set \(x=-1\). The \(B\)-term vanishes, so \(2=A(1)\), giving \(A=2\). Set \(x=-2\). The \(A\)-term vanishes, so \(-1=B(-1)\), giving \(B=1\). Thus
Recombining confirms the identity: \(2(x+2)+(x+1)=3x+5\). Integrate each term on any interval that avoids \(-1\) and \(-2\):
Differentiate the result to check it:
Why the Decomposition Works
The coefficient-finding steps rely on a genuine algebraic fact: every proper rational function with a denominator made of distinct linear factors has exactly one decomposition of the stated form. Here is a proof that also gives a formula for the coefficients.
To prove the claim, multiply the proposed identity by \(\prod_{i=1}^{n}(x-a_i)\). The required polynomial identity is
Set \(x=a_i\). Every term on the right vanishes except the term with coefficient \(A_i\). Therefore, if the identity holds,
The factors \(a_i-a_j\) are nonzero because the \(a_i\) are distinct. Dividing gives the stated formula for \(A_i\), so any decomposition must have these coefficients; this proves uniqueness. To prove existence, define each \(A_i\) by that formula and let the difference between the two sides of the polynomial identity be \(R(x)\). Its degree is at most \(n-1\). The formula ensures \(R(a_i)=0\) for all \(n\) distinct values \(a_i\). A polynomial of degree at most \(n-1\) with \(n\) distinct roots must be the zero polynomial. Thus the identity holds, proving existence as well.
In practice, multiplying through and evaluating at each \(a_i\) is often the quickest way to obtain the coefficients. The formula also explains why these substitutions work: at a factor’s zero, the other terms in the cleared identity disappear.
Integrating the Decomposed Terms
For a real number \(a\), the derivative of \(\ln|x-a|\) is \(1/(x-a)\) whenever \(x\ne a\). Therefore, if a decomposition has been established and the interval of integration contains none of its denominator zeros, then
The absolute value matters: it makes the logarithm defined on either side of \(a_i\), as long as \(x\ne a_i\). An antiderivative formula applies separately on each interval that avoids all the zeros. The constant of integration can differ between those intervals.
Worked Example: Three Distinct Factors
Decompose and integrate \(\frac{4x^2+5x+3}{(x-1)(x+1)(x+2)}\).
The numerator has degree \(2\), less than the denominator’s degree \(3\). Write
Clearing denominators gives
Substitute each denominator zero. At \(x=1\), the equation is \(12=6A\), so \(A=2\). At \(x=-1\), it is \(2=-2B\), so \(B=-1\). At \(x=-2\), it is \(9=3C\), so \(C=3\). Hence
Check by recombining: the numerator is \(2(x+1)(x+2)-(x-1)(x+2)+3(x-1)(x+1)\). Expanding gives \(2x^2+6x+4-x^2-x+2+3x^2-3=4x^2+5x+3\). Integrating on an interval that avoids \(-2\), \(-1\), and \(1\) gives
Differentiate to check: the derivative is \(\frac{2}{x-1}-\frac{1}{x+1}+\frac{3}{x+2}\). Recombining these terms gives \(\frac{4x^2+5x+3}{(x-1)(x+1)(x+2)}\), as required.
Using the Decomposition in a Definite Integral
A definite integral can use the same process, provided the interval contains no denominator zero. In that case the integrand is continuous throughout the interval, so its antiderivative can be evaluated at the endpoints.
Worked Example: Evaluate a Definite Integral
Evaluate \(\int_0^1 \frac{1}{(x+1)(x+3)}\,dx\).
The denominator has two distinct factors and the numerator degree is less than the denominator degree. Write
Clearing denominators gives \(1=A(x+3)+B(x+1)\). Setting \(x=-1\) gives \(1=2A\), so \(A=\frac12\). Setting \(x=-3\) gives \(1=-2B\), so \(B=-\frac12\). Thus
There are no denominator zeros between \(0\) and \(1\). An antiderivative is \(\frac12\ln|x+1|-\frac12\ln|x+3|\). Evaluate at the endpoints:
To check the antiderivative, differentiate it: \(\frac{1}{2(x+1)}-\frac{1}{2(x+3)}=\frac{(x+3)-(x+1)}{2(x+1)(x+3)}=\frac{1}{(x+1)(x+3)}\).
Common Mistakes and Checks
A frequent error is to guess the coefficients by matching the numerator’s coefficients directly to the partial-fraction numerators. That is not how the identity works: clearing denominators multiplies each coefficient by all the other factors.
For example, in the first worked example, a wrong guess is \(\frac{3}{x+1}+\frac{5}{x+2}\), based on the numerator \(3x+5\). Recombining shows the error:
which is not the original fraction. The correct constants are \(2\) and \(1\), because clearing denominators gives \(3x+5=A(x+2)+B(x+1)\); substituting \(x=-1\) and \(x=-2\) yields those values. Recombining is a useful check whenever coefficients are found.
Also avoid dropping absolute value bars from logarithms in a general antiderivative. The expression \(\ln(x-a)\) is defined only for \(x>a\), while \(\ln|x-a|\) works on either side of \(a\), excluding the point \(a\) itself. Finally, do not apply this setup to a fraction that is not proper without first using polynomial division, or to a denominator that does not consist of distinct linear factors; those situations require a different setup.
Check Your Understanding
Try these questions. For each decomposition, state any values of \(x\) where the original function is undefined.
- For \(\frac{2x+7}{(x+1)(x+3)}\), write the partial-fraction form and determine its constants.
- Decompose \(\frac{x^2+2x+4}{x(x+1)(x+2)}\), then write an antiderivative.
- Why must the numerator degree be less than the denominator degree before using this form directly?
- Why does evaluating the cleared identity at a denominator zero isolate one coefficient?
- What is the role of the absolute value in \(\ln|x-a|\), and where can the antiderivative be used?
