Why the Discriminant Matters in Integration
A quadratic denominator can lead to different integration methods depending on whether it has distinct real roots, a repeated real root, or no real roots. The discriminant makes that distinction quickly. Completing the square then displays the same information in a form that also helps identify an appropriate integration route.
This classification is a decision-making tool, not a method for integrating every possible numerator in one step. For a rational function, the numerator still matters. The denominator’s factorization tells us which kinds of terms may arise and which approach is suitable.
Definition and Completed-Square Form
Completing the square gives a useful identity. Set \(h=-\frac{b}{2a}\). Then
Thus the sign of \(\Delta\) determines whether the squared term must equal a positive number, zero, or a negative number for \(Q(x)=0\). This leads to the three cases below.
| Discriminant | Roots over the real numbers | Factorization over the real numbers |
|---|---|---|
| \(\Delta>0\) | Two distinct real roots | Two distinct linear factors |
| \(\Delta=0\) | One repeated real root | A repeated linear factor |
| \(\Delta<0\) | No real roots | Irreducible as a product of real linear factors |
When \(\Delta<0\), the completed-square expression can also be written as
The quantity added to the square is strictly positive. Therefore \(Q(x)\) never vanishes for real \(x\); its sign is the sign of \(a\). “Irreducible” here means that \(Q\) cannot be factored into real linear factors, not that it cannot be expressed using complex roots.
Why the Three Cases Follow
The completed-square identity provides a direct proof of the classification. For a root, \(Q(x)=0\), so the identity gives \((x-h)^2=\frac{\Delta}{4a^2}\). A square of a real number is nonnegative, and \(4a^2\) is positive because \(a\ne0\). The sign of \(\Delta\) therefore determines the possibilities.
The equation becomes \((x-h)^2=\frac{\Delta}{4a^2}>0\). It has the two solutions \(x=h\pm\frac{\sqrt{\Delta}}{2|a|}\), which are distinct real numbers.
The equation becomes \((x-h)^2=0\), so its only solution is \(x=h\). The factorization is \(Q(x)=a(x-h)^2\), showing that this root is repeated.
The equation would require a nonnegative square to equal a negative number, which is impossible over the real numbers. Hence there are no real roots.
This also explains why the discriminant is useful when integrating: it reveals the real factor structure of the denominator before any decomposition or substitution is attempted.
Choosing an Integration Approach
Consider first an integral with a quadratic denominator and a rational numerator. If \(\Delta>0\), the denominator can be factored into distinct real linear factors. For a reciprocal of that denominator, partial fractions express the integrand as a sum of terms with linear denominators. If \(\Delta=0\), the denominator is a repeated linear factor, so the decomposition must account for its repeated power. These structures often lead to logarithmic terms or reciprocal powers, depending on the numerator and the resulting terms.
If \(\Delta<0\), there are no real linear factors to use in a real partial-fraction decomposition. Completing the square writes the denominator as a constant times a positive quadratic expression of the form \((x-h)^2+d^2\), with \(d>0\). A reciprocal of this expression is associated with an inverse-tangent form. A numerator proportional to the derivative of the denominator, on the other hand, can lead to a logarithm. The numerator and the completed-square form together determine which route is appropriate.
Worked Examples
Worked Example: Two Distinct Real Roots
Classify \(Q(x)=x^2-5x+6\), then integrate its reciprocal on an interval that contains neither root.
Here \(a=1\), \(b=-5\), and \(c=6\). The discriminant is
There are two distinct real roots. Factoring gives \(Q(x)=(x-2)(x-3)\), since \(2+3=5\) and \(2\cdot3=6\). Use partial fractions and solve for the constants:
Setting \(x=2\) gives \(1=-A\), so \(A=-1\). Setting \(x=3\) gives \(1=B\), so \(B=1\). Therefore, on any interval not containing \(2\) or \(3\),
Check by differentiating. The derivative is
This recovers the original integrand. The antiderivative is valid separately on intervals that do not cross either point where the integrand is undefined.
Worked Example: A Repeated Real Root
Classify \(Q(x)=4x^2+4x+1\), then integrate its reciprocal wherever it is defined.
Here \(a=4\), \(b=4\), and \(c=1\). Thus
The root is repeated. Factoring, or using the completed square, gives
To integrate, set \(u=2x+1\), so \(du=2\,dx\) and \(dx=\frac{du}{2}\). Then
Check by differentiating the proposed antiderivative:
This is the original integrand. The function is undefined at \(x=-\frac12\), so the antiderivative applies on intervals that do not include that point.
Worked Example: No Real Roots and a Numerator That Matches the Derivative
Classify \(Q(x)=x^2+2x+5\), then integrate \(\frac{2x+2}{Q(x)}\).
Here \(a=1\), \(b=2\), and \(c=5\), so
There are no real roots. Completing the square gives
Since \(Q'(x)=2x+2\), the numerator is exactly the derivative of the denominator. With \(u=Q(x)\), \(du=(2x+2)\,dx\), so
The absolute value can be removed because \((x+1)^2+4>0\) for every real \(x\). Check:
which is the original integrand. For comparison, if the numerator were \(1\) instead, completing the square and setting \(u=\frac{x+1}{2}\) would give
This is the inverse-tangent type of integral. The same irreducible denominator can therefore call for different approaches depending on its numerator.
A Common Sign Error
A frequent mistake is to confuse \(b^2-4ac\) with \(4ac-b^2\). For \(x^2+4x+5\), the discriminant is
A wrong calculation might use \(4ac-b^2=20-16=4\) and conclude that the quadratic has two real roots. That quantity is the negative of the discriminant, so the conclusion is incorrect. The correction is to use the stated definition consistently: \(\Delta=b^2-4ac=-4<0\). Indeed, completing the square gives \(x^2+4x+5=(x+2)^2+1\), which is positive for all real \(x\) and has no real roots.
Using the Classification Carefully
The discriminant is most useful as an early diagnostic. First compute it, then choose a route compatible with the resulting factor structure. For a general rational integrand, do not assume that a denominator classification alone settles the whole problem: a numerator may simplify the integral, match the derivative of the denominator, or require an additional algebraic step.
Also keep track of where the original integrand is defined. A real root of the denominator gives a point that must be excluded, and antiderivatives are considered on intervals that do not cross such points. In the no-real-root case, the quadratic denominator has no real zeros, so this particular source of domain restrictions is absent.
Check Your Understanding
For each question, use the discriminant and, where useful, the completed-square form to justify your answer.
- Classify \(3x^2-6x+3\) by its discriminant. What is its real factorization?
- For \(x^2-2x+5\), calculate the discriminant and write the completed-square form. Can its denominator vanish for a real \(x\)?
- What integration approach is suggested for the reciprocal of a quadratic with two distinct real roots?
- Why does a repeated real root require attention to the power of the linear factor in an integration method?
- For an irreducible quadratic, why should you inspect the numerator before choosing an integration approach?
