Rational Definite Integrals By Substitution

From a Composite Denominator to a Logarithm

A rational integrand has a polynomial expression divided by another polynomial expression. Some such integrals become much simpler when the numerator is the derivative of the denominator, or a constant multiple of that derivative. The key pattern is a function divided by its own derivative’s source: \(g'(x)/g(x)\).

This pattern leads to a logarithm because the derivative of a logarithm of a nonzero function is its derivative divided by the function. For a definite integral, substitution also changes the integration limits. Transforming those limits as well as the integrand keeps the calculation entirely in the new variable.

Recognizing the Matching Factors

Suppose a definite integral has the form

where \(c\) is a constant, \(g\) is differentiable on \([a,b]\), and \(g(x)\ne 0\) at every point of that interval. The denominator is the proposed inner function, and its derivative should match the numerator apart from the constant \(c\).

For example, if the denominator is \(3x^2+2x+5\), its derivative is \(6x+2\). A numerator of \(6x+2\) matches exactly. A numerator of \(9x+3\) is \(\frac{3}{2}(6x+2)\), so it matches with constant factor \(\frac{3}{2}\).

Matching is a structural check, not a guess based only on how complicated the denominator looks. Differentiate the denominator explicitly, then compare every term with the numerator. If the numerator does not match, this particular logarithmic pattern does not directly evaluate the whole integral.

Substitution and the Changed Bounds

Set \(u=g(x)\). Then \(du=g'(x)\,dx\), so a numerator of \(c\,g'(x)\) becomes \(c\,du\). The original limits \(x=a\) and \(x=b\) must also be converted: the new lower limit is \(g(a)\), and the new upper limit is \(g(b)\).

The logarithm in the antiderivative is \(\ln|u|\), since the natural logarithm is defined for positive inputs and \(u\) need not be positive. The condition \(g(x)\ne 0\) throughout the interval ensures that the original integrand has no denominator singularity there and that the logarithmic expression is defined along the interval.

Why the Logarithmic Formula Works

The formula follows from differentiation and the Fundamental Theorem of Calculus. Because \(g\) is continuous and never zero on the interval, it cannot change sign there: a sign change would force it to equal zero at some point. Thus either \(g(x)>0\) throughout or \(g(x)<0\) throughout.

If \(g\) is positive, the chain rule gives \(\frac{d}{dx}\ln(g(x))=\frac{g'(x)}{g(x)}\). If \(g\) is negative, then \(|g(x)|=-g(x)\), so

In either case, \(\ln|g(x)|\) is an antiderivative of \(g'(x)/g(x)\) on the interval. The Fundamental Theorem of Calculus therefore gives the definite-integral formula. Multiplying the integrand by a constant \(c\) multiplies the integral by \(c\).

Worked Examples

A Common Bounds Mistake

Once the integrand has been rewritten in terms of \(u\), the integral’s bounds must also be expressed in terms of \(u\). Keeping the old \(x\)-bounds while integrating with respect to \(u\) mixes two variables and generally produces the wrong value.

For the first example, \(u=3x^2+2x+5\). A mistaken calculation would leave the limits as \(0\) and \(2\):

This is not a valid evaluation: \(\ln(0)\) is undefined, and \(0\) and \(2\) were values of \(x\), not the corresponding values of \(u\). The correct transformed bounds are \(u=5\) and \(u=21\), giving \(\int_5^{21}\frac{du}{u}=\ln(21)-\ln(5)\).

A separate condition matters as well: the denominator must not vanish anywhere in the interval for this ordinary definite-integral formula to apply. If it does vanish, the integrand is not continuous there, and the integral requires separate treatment rather than direct substitution across the singularity.

Using the Result Reliably

For a rational definite integral with a composite denominator, the practical test is to differentiate the denominator and compare the result with the numerator. When they match up to a constant and the denominator stays nonzero on the interval, substitution reduces the calculation to the integral of \(1/u\).

The transformed limits encode the original endpoints. They also make the final logarithm a comparison of the denominator’s endpoint values. In particular, the answer depends on both endpoint values, not just on the width of the original interval. A final differentiation check confirms that the proposed logarithmic expression has the required integrand as its derivative.