From a Mixed Denominator to an Integral
A rational function can be difficult to integrate in its original form even when its denominator factors into familiar pieces. Partial fractions replace it with a sum of simpler terms, each matched to one denominator factor. When a linear factor is repeated and an irreducible quadratic factor is also present, the decomposition must include terms for every power of each factor.
Here we work with proper rational functions: the numerator has lower degree than the denominator. We factor the denominator over the real numbers, construct the required decomposition, determine its coefficients, and integrate term by term. We will check both the decomposition and the resulting antiderivative.
The numerator over a quadratic factor is linear, not just constant: its degree must be less than the degree of the quadratic. The decomposition’s coefficients are constants, and some may be zero. After finding them, integrate the terms separately. Linear factors produce logarithms or reciprocal powers; quadratic terms may produce logarithms and arctangents, depending on their numerators.
Why Clearing the Denominator Works
Suppose a proposed decomposition is an identity of rational functions wherever the denominators are nonzero. Multiplying both sides by their common denominator gives a polynomial equation. Conversely, if that polynomial equation holds, dividing by the common denominator recovers the rational-function identity at every point where the original denominator is nonzero.
This gives a practical coefficient-solving method: clear denominators, expand, and match coefficients of equal powers of \(x\). Substituting a root of a factor can also isolate a coefficient. In particular, if the linear factor is repeated, substitution at its root isolates the coefficient of the highest reciprocal power. These are ways of solving the same polynomial identity, not extra assumptions about the function.
Worked Decomposition and Integration
Worked Example: A Repeated Linear Factor and a Quadratic
Integrate the proper rational function
The numerator has degree \(3\), while the denominator has degree \(4\), so the function is proper. Since \(x^2+1\) has no real roots, use the form
Multiply by \((x-1)^2(x^2+1)\):
Set \(x=1\). The terms with \(A,C,D\) vanish, giving \(-2=2B\), so \(B=-1\). Expanding the remaining identity gives the coefficient equations
Using \(B=-1\), the constant equation gives \(-A-1+D=1\), hence \(D-A=2\). The cubic equation gives \(C=5-A\). Substitution in the linear equation gives \(A+(5-A)-2D=-3\), so \(D=4\), then \(A=2\) and \(C=3\). The quadratic equation checks: \(-2-1+4-6=-5\).
Thus the decomposition is
Integrate term by term. In the last term, split the numerator as \(3x+4=3x+4\): the \(3x\) part gives a logarithm, and the constant part gives an arctangent.
Therefore an antiderivative is
Check the decomposition: Combining its numerators over the common denominator gives \(2(x-1)(x^2+1)-(x^2+1)+(3x+4)(x-1)^2\). Expanding gives \(5x^3-5x^2-3x+1\), as required.
Check the antiderivative: Differentiating \(F\) gives \(2/(x-1)-1/(x-1)^2+3x/(x^2+1)+4/(x^2+1)\), exactly the decomposition.
A Shifted Quadratic Factor
Worked Example: Using a Shifted Variable
Integrate
The denominator has degree \(4\), greater than the numerator’s degree \(3\). Set \(t=x+2\); then \(x=t-2\) and \(x^2+4x+5=t^2+1\). Substitution in the numerator gives \(6(t-2)^3+38(t-2)^2+82(t-2)+63=6t^3+2t^2+2t+3\). The decomposition in \(t\) has the form
Clearing the denominator gives \(6t^3+2t^2+2t+3=A t(t^2+1)+B(t^2+1)+(Ct+E)t^2\). Matching the coefficients of \(t^3,t^2,t,1\), in that order, gives \(A+C=6,\ B+E=2,\ A=2,\ B=3\). Thus \(A=2,\ B=3,\ C=4,\ E=-1\). Returning to \(x\),
For the quadratic term, write \(4x+7=2(2x+4)-1\), where \(2x+4\) is the derivative of \(x^2+4x+5\). Also \(x^2+4x+5=(x+2)^2+1\). Hence
The complete antiderivative is
Check the decomposition: In the variable \(t\), recombination gives \(2t(t^2+1)+3(t^2+1)+(4t-1)t^2=6t^3+2t^2+2t+3\). This is the transformed numerator, so the identity holds.
Check the antiderivative: Its derivative is \(2/(x+2)+3/(x+2)^2+(4x+7)/(x^2+4x+5)\). For the reciprocal term, \(\frac{d}{dx}[-3/(x+2)]=3/(x+2)^2\). The derivative matches the decomposition.
Adding a Simple Linear Factor
Worked Example: Three Types of Factors
Consider the integral
The denominator has degree \(5\), and the numerator has degree \(4\). The denominator contains a repeated linear factor, a simple linear factor, and an irreducible quadratic. Use
After clearing denominators,
where \(P(x)=6x^4-9x^3+x^2-7x+1\). At \(x=1\), \(P(1)=-8=4B\), so \(B=-2\). At \(x=-1\), \(P(-1)=24=8C\), so \(C=3\).
For the remaining coefficients, expand the terms involving \(A,D,E\), together with the known \(B,C\). Coefficient comparison gives
The last two equations give \(D=2\) and \(E=1\); then \(A=1\). The constant coefficient also checks: \(-A+B+C+E=-1-2+3+1=1\). Therefore
Integrating each term gives
Check the decomposition: Recombining the numerators produces \((x^4-1)-2(x^3+x^2+x+1)+3(x^4-2x^3+2x^2-2x+1)+(2x+1)(x^3-x^2-x+1)\). Adding like powers gives \(6x^4-9x^3+x^2-7x+1\).
Check the antiderivative: Differentiation gives \(1/(x-1)-2/(x-1)^2+3/(x+1)+(2x+1)/(x^2+1)\), the displayed decomposition.
Integration Rules and Their Conditions
The examples use standard antiderivatives. The logarithm rule applies to a nonzero linear expression on an interval where it does not vanish. For the positive quadratic \(x^2+1\), the logarithm and arctangent expressions are defined for all real \(x\). More generally, the domain of the original rational function excludes every real zero of its denominator, so an antiderivative is considered on an interval that does not cross such a zero.
In a quadratic numerator, first separate a multiple of the derivative of the denominator from the remaining constant. For example, in \(4x+7\) over \(x^2+4x+5\), the derivative is \(2x+4\), and \(4x+7=2(2x+4)-1\). This separation identifies the logarithmic part and leaves a constant-over-quadratic part for the arctangent.
Common Pitfall: Omitting a Repeated-Factor Term
A repeated factor requires terms for each of its powers. For \((x-1)^2\), writing only \(A/(x-1)\) is incomplete. In the first example, that omission would require the entire numerator to be expressible as \(A(x-1)(x^2+1)+(Cx+D)(x-1)^2\). Every term on the right is zero at \(x=1\), but the actual numerator at \(x=1\) is \(-2\). The proposed identity cannot hold.
The missing term \(B/(x-1)^2\) contributes \(B(x^2+1)\) after clearing denominators, which need not vanish at \(x=1\). The substitution then gives \(2B=-2\), so \(B=-1\). This concrete check explains why every power in the repeated factor must have its own term.
Why the Decomposition Is Useful
The decomposition turns one complicated fraction into terms whose antiderivatives are recognizable. Just as importantly, the polynomial identity gives a reliable check: recombination must recover the original numerator exactly. Differentiating the final expression gives a second, independent check that the integration was carried out correctly.
Check Your Understanding
Use the factor structure and checks in this tutorial to answer the following questions.
- What terms belong in the decomposition for a denominator containing \((x-3)^2(x^2+4)^2\)?
- Why does multiplying a proposed decomposition by the common denominator produce a polynomial identity?
- For \(x^2+4x+5\), what is its derivative, and how can that help integrate a linear numerator over it?
- Why is a decomposition for a denominator containing \((x-1)^2\) incomplete if it includes only \(A/(x-1)\)?
- What two checks can be used after finding an antiderivative by partial fractions?
