Recognizing the Matching Structure
A rational function with a quadratic denominator may have a numerator that looks complicated at first. But if its linear numerator is proportional to the derivative of the quadratic, the integral reduces to a direct substitution. This works whether the denominator is a first power or a higher positive integer power.
The key is to compare the numerator with the derivative of the entire quadratic expression in the denominator. If they match up to a constant factor, substitution turns the integral into a logarithm or a simple power integral.
The constant \(\lambda\) matters: it is the factor that remains after substitution. Matching does not require the numerator and derivative to be exactly equal, only proportional.
The Substitution Rule
Consider an integral whose denominator is \(Q(x)^n\), with \(n\) a positive integer, and whose numerator is \(kQ'(x)\), with \(k\) constant. Set \(u=Q(x)\). Then \(du=Q'(x)\,dx\), so the integral becomes \(k\int u^{-n}\,du\). For \(n=1\), this gives a logarithm; for \(n>1\), the ordinary power rule applies.
The condition \(Q(x)\ne 0\) is necessary because the original integrand is undefined where its denominator is zero. The absolute value in the logarithm allows \(Q(x)\) to be either positive or negative on an interval that does not cross a zero. In the common case of an irreducible quadratic that is positive for every real \(x\), the absolute value can be omitted, but the general formula uses it.
Why the Rule Works
The substitution is not a trick specific to quadratics. It works because the differential \(Q'(x)\,dx\) is present in the numerator. Here is the complete derivation for the higher-power case.
Set \(u=Q(x)\). Then \(du=Q'(x)\,dx\), so the numerator factor \(Q'(x)\,dx\) is replaced by \(du\).
For \(n>1\), the integral becomes \(k\int u^{-n}\,du\).
Since \(n\ne 1\), an antiderivative of \(u^{-n}\) is \(u^{1-n}/(1-n)\).
Multiplying by \(k\) and replacing \(u\) with \(Q(x)\) gives \(-k/((n-1)Q(x)^{n-1})+C\).
The result can also be checked directly by differentiation. The chain rule gives
For the first-power case, the chain rule gives
where \(Q(x)\ne 0\). Thus both formulas differentiate back to their integrands.
Worked Examples
Worked Example: A Derivative-Matched Numerator Over a First Power
Evaluate \(\int \frac{10x-6}{5x^2-6x+9}\,dx\).
Let \(Q(x)=5x^2-6x+9\). Its derivative is
The numerator is exactly \(Q'(x)\), so \(k=1\). The denominator is a first power, and therefore
To check, differentiate the proposed antiderivative:
This matches the original integrand. Also, \(5x^2-6x+9\) is positive for every real \(x\): its discriminant is \((-6)^2-4(5)(9)=36-180=-144<0\), and its leading coefficient is positive. Thus the absolute value could be omitted in this example, though it is valid to keep it.
Worked Example: A Proportional Numerator Over a Higher Power
Evaluate \(\int \frac{8x+12}{(2x^2+6x+11)^2}\,dx\).
Set \(Q(x)=2x^2+6x+11\). Then
The numerator is \(2Q'(x)\), so \(k=2\), and the denominator has power \(n=2\). The higher-power formula gives
Differentiate to check:
The derivative is the original integrand, including its factor of \(2\).
Worked Example: A Definite Integral with a Nonunit Proportionality Factor
Evaluate \(\int_0^1 \frac{3x+6}{(x^2+4x+8)^3}\,dx\).
Take \(Q(x)=x^2+4x+8\). Then \(Q'(x)=2x+4\), and
Use \(u=Q(x)\), so \(du=Q'(x)\,dx\). The limits change as follows:
Therefore,
For clarity, \(676\cdot256=173{,}056\), so the preceding fraction is \(1260/173{,}056\). Dividing numerator and denominator by \(4\) gives \(315/43{,}264\). The definite integral is thus \(315/43{,}264\).
A check comes from the antiderivative \(-3/(4Q(x)^2)\). Its derivative is
Evaluating this antiderivative at \(1\) and \(0\) gives \(-3/(4\cdot169)+3/(4\cdot64)\), which is the value computed above.
Checking Whether a Numerator Matches
For a quadratic \(Q(x)=ax^2+bx+c\), the derivative is \(2ax+b\). Suppose the numerator is \(mx+r\). To test for a match, compare coefficients in
The coefficients must satisfy \(m=2a\lambda\) and \(r=b\lambda\) for the same constant \(\lambda\). If \(m\ne 0\), one can compute \(\lambda=m/(2a)\) and check whether \(r=bm/(2a)\). Matching only the coefficient of \(x\) is not enough: the constant terms must match as well.
For example, with \(Q(x)=3x^2+5x+4\), the derivative is \(6x+5\). A numerator \(12x+10\) matches because it is \(2Q'(x)\). A numerator \(12x+9\) does not: twice the derivative would be \(12x+10\), not \(12x+9\). The direct substitution rule applies to the matching part of a numerator, not automatically to an entire numerator that fails this test.
A Common Mistake: Losing the Proportionality Factor
A frequent error is to notice that the numerator is proportional to \(Q'(x)\), but then proceed as if the proportionality factor were \(1\). For instance, in the second worked example, the numerator is \(2Q'(x)\). Omitting that factor would give the incorrect answer \(-1/Q(x)+C\).
Differentiate that incorrect answer:
which is only half the required integrand \(2Q'(x)/Q(x)^2\). The correction is to carry \(k=2\) through the integration:
Differentiating this corrected answer gives \(2Q'(x)/Q(x)^2\), as required. Checking by differentiation is an especially effective way to catch missing constants and incorrect powers.
When This Method Is Useful
Derivative-matched terms commonly appear after a rational function has been decomposed into partial fractions involving repeated irreducible quadratic factors. A term with a linear numerator may integrate immediately if that numerator is proportional to the derivative of the quadratic. Recognizing this structure can turn an expression that appears to need several steps into one substitution.
The method is deliberately specific. It handles a numerator proportional to the derivative of the denominator’s quadratic base. It does not by itself integrate every linear numerator over a quadratic power. If the numerator does not match, first identify exactly which part, if any, is derivative-matched; the remaining part requires a different integration step. The key practical sequence is to differentiate the quadratic, compare both numerator coefficients, retain the proportionality constant, and then verify the antiderivative.
Check Your Understanding
For each question, identify the quadratic and its derivative before choosing an integration rule.
- Evaluate \(\int \frac{4x+2}{x^2+x+5}\,dx\).
- Evaluate \(\int \frac{6x+9}{(3x^2+9x+12)^3}\,dx\).
- For \(Q(x)=2x^2+7x+1\), is \(10x+17\) proportional to \(Q'(x)\)? Explain by comparing coefficients.
- Why does the formula for a first power involve \(\ln|Q(x)|\), while the formula for powers \(n>1\) has a denominator \(Q(x)^{n-1}\)?
- Differentiate \(-5/(4Q(x)^4)\), where \(Q(x)=x^2+3x+6\), and state the resulting integrand.
