Why Irreducible Quadratics Need a Different Numerator
A rational function can have denominator factors that do not split into real linear factors. When that happens, its partial-fraction decomposition must account for those factors directly. The key is that the numerator over an irreducible quadratic is linear, not just a constant.
After decomposing the rational function, each term over a quadratic can be integrated by separating the part related to the derivative of the denominator from the remaining constant part. These two pieces lead to different familiar antiderivatives: logarithms and inverse tangents.
An irreducible quadratic has no real zeros, so it does not create real vertical asymptotes. For partial fractions, the main consequence is structural: a term over an irreducible quadratic has the form \(\frac{Ax+B}{q(x)}\).
The Form of the Decomposition
Suppose a proper rational function has a denominator that factors into distinct real linear factors and distinct irreducible quadratic factors. A constant numerator is used over each linear factor. A linear numerator is used over each irreducible quadratic factor. For example, if the denominator is \((x-1)(x^2+1)(x^2+4)\), the decomposition has the form
Here \(P(x)\) is the numerator, and \(A,B,C,D,E\) are constants to be determined. The denominator factors in this form are distinct; repeated irreducible quadratic factors require a different setup.
The degree rule explains why the quadratic numerator must be linear. When dividing a polynomial by a quadratic, the remainder has degree less than two, so it has the form \(Bx+C\). Likewise, the remainder over a linear factor is a constant. This is the reason for the shapes of the partial-fraction terms.
Integrating a Linear Numerator over a Quadratic
Consider a term \(\frac{Ax+B}{ax^2+bx+c}\), where \(a\ne0\) and \(b^2-4ac<0\). The derivative of the denominator is \(2ax+b\). We can express the numerator as a multiple of that derivative plus a constant:
The first piece integrates to a logarithm. To integrate the constant piece, complete the square in the quadratic. The result is the following formula.
Because \(q\) has no real zeros, its sign is constant on the real line, so \(\ln|q(x)|\) is defined for every real \(x\). Including the absolute value also keeps the logarithm expression correct when \(q(x)\) is negative.
Why the Formula Works
First split the numerator using the derivative \(q'(x)=2ax+b\). Then complete the square. Since \(\Delta=4ac-b^2>0\),
Consequently, differentiating the logarithm gives the derivative-related part of the numerator, while completing the square puts the remaining denominator into the standard inverse-tangent form.
Write \(Ax+B=\frac{A}{2a}q'(x)+\left(B-\frac{Ab}{2a}\right)\).
The chain rule gives \(\int \frac{q'(x)}{q(x)}\,dx=\ln|q(x)|+C\).
Use the substitution \(u=\frac{2ax+b}{\sqrt{\Delta}}\) to obtain an inverse tangent.
For the last step, the substitution has \(du=\frac{2a}{\sqrt{\Delta}}\,dx\). Since \(q(x)=\frac{(2ax+b)^2+\Delta}{4a}\), the remaining integral satisfies
Multiplying this by the constant left after splitting the numerator gives the inverse-tangent coefficient in the theorem. Together with the logarithm term, this proves the formula.
Worked Example: Integrating Both Numerator Parts
Worked Example: A Quadratic with a Simple Derivative
Evaluate \(\int\frac{2x+3}{x^2+4}\,dx\). The denominator has derivative \(2x\), so split the numerator into \(2x+3\):
For the first integral, let \(u=x^2+4\), so \(du=2x\,dx\). This gives \(\int\frac{2x}{x^2+4}\,dx=\ln(x^2+4)\). For the second, use \(\int\frac{dx}{x^2+a^2}=\frac{1}{a}\arctan(\frac{x}{a})+C\), with \(a=2\). Therefore,
Check by differentiating. The derivative of the logarithm is \(\frac{2x}{x^2+4}\). The derivative of the inverse-tangent term is \(\frac{3}{2}\cdot\frac{1/2}{1+x^2/4}=\frac{3}{x^2+4}\). Adding gives \(\frac{2x+3}{x^2+4}\), as required.
Worked Example: Decomposing Two Distinct Quadratics
Worked Example: Find the Decomposition, Then Integrate
Evaluate
The denominator contains two distinct irreducible quadratics. Set up the decomposition with a linear numerator over each:
Multiply both sides by \((x^2+1)(x^2+4)\):
Expanding the right-hand side gives \((A+C)x^3+(B+D)x^2+(4A+C)x+(4B+D)\). Equating coefficients gives \(A+C=3\), \(B+D=-1\), \(4A+C=9\), and \(4B+D=2\). Subtracting the first equation from the third gives \(3A=6\), so \(A=2\) and \(C=1\). Subtracting the second equation from the fourth gives \(3B=3\), so \(B=1\) and \(D=-2\). Thus,
Now integrate each term. In the first fraction, the numerator is \(2x+1\), so separate \(2x\) and \(1\). In the second, separate \(x\) and \(-2\):
The antiderivative is therefore
To check, differentiation gives \(\frac{2x+1}{x^2+1}+\frac{x-2}{x^2+4}\). Recombining those fractions gives numerator \((2x+1)(x^2+4)+(x-2)(x^2+1)=3x^3-x^2+9x+2\), over \((x^2+1)(x^2+4)\). This is the original integrand.
Worked Example: Completing the Square in a Shifted Quadratic
Worked Example: A Shifted Denominator
Evaluate \(\int\frac{4x+6}{x^2+2x+5}\,dx\). The denominator’s derivative is \(2x+2\), and the numerator can be written as \(2(2x+2)+2\). Therefore,
The first integral is \(2\ln(x^2+2x+5)\). For the second, complete the square: \(x^2+2x+5=(x+1)^2+4\). Thus,
The answer is \(2\ln(x^2+2x+5)+\arctan(\frac{x+1}{2})+C\). Check: the derivative of the logarithm term is \(\frac{4x+4}{x^2+2x+5}\). The derivative of the inverse-tangent term is \(\frac{1/2}{1+(x+1)^2/4}=\frac{2}{x^2+2x+5}\). Their sum is \(\frac{4x+6}{x^2+2x+5}\), the original integrand.
A Common Mistake: Forgetting the Derivative Factor
A frequent error is to treat the coefficient in front of a derivative-related numerator as if it remains unchanged in the logarithm. For instance, the incorrect calculation
has an extra factor of \(2\). Differentiating its right-hand side produces \(\frac{4x}{x^2+4}\), not \(\frac{2x}{x^2+4}\). The correction is to notice that the numerator \(2x\) is exactly the derivative of \(x^2+4\):
More generally, if the numerator is a multiple \(kq'(x)\), then \(\int\frac{kq'(x)}{q(x)}\,dx=k\ln|q(x)|+C\). The coefficient \(k\) comes from writing the numerator as a multiple of the full derivative, not from an extra multiplication after integrating.
Putting the Method to Work
For a proper rational function with distinct irreducible quadratic factors, first write the appropriate linear numerator over each quadratic. Solve for the constants by clearing denominators and comparing coefficients. Then integrate each resulting term by splitting its numerator into a multiple of the denominator’s derivative plus a constant. Check the final result by differentiating or, before integration, recombining the decomposition.
This method matters because recognizing the derivative-related part makes logarithmic terms visible, while completing the square reveals the inverse-tangent term. Neither part should be omitted: a linear numerator can contribute both.
Check Your Understanding
For each question, identify the relevant decomposition or integration step and show enough work to justify your answer.
- What form should the partial-fraction term over \(x^2+9\) have?
- Find \(\int\frac{2x+5}{x^2+9}\,dx\).
- Set up a partial-fraction decomposition for a proper rational function with denominator \((x^2+1)(x^2+4)\).
- Complete the square in \(x^2-4x+8\), then state the inverse-tangent form associated with its reciprocal.
- How can differentiating an antiderivative reveal an incorrect coefficient on a logarithm term?
