Recognizing the Denominator Pattern
A rational integrand is a quotient of expressions. A particularly direct case occurs when the numerator is the derivative of the denominator, perhaps multiplied by a constant. The denominator may itself be a composite expression, such as a polynomial raised to a power.
This pattern is useful because it reverses the chain rule. Instead of expanding or otherwise rearranging the integrand, we take the entire denominator's inner expression as the substitution variable. The resulting integral is either a logarithm or a simple power.
The key recognition step is to compare the numerator with the derivative of the whole denominator's inner expression. For instance, if the denominator is \((x^2+4)^3\), the expression to differentiate is \(x^2+4\), not the entire power. Its derivative is \(2x\). A numerator proportional to \(2x\) matches; a numerator that does not match this way is not an instance of this direct pattern.
The Substitution and Its Formulas
Let \(u=g(x)\). Then \(du=g'(x)\,dx\). If the numerator is \(k g'(x)\), the integral becomes \(k\int u^{-n}\,du\). When \(n=1\), the antiderivative is logarithmic. When \(n>1\), the ordinary power rule applies to \(u^{-n}\).
The condition \(g(x)\ne0\) matters: the original integrand has a denominator, so it is undefined wherever that denominator is zero. Antiderivatives are considered on intervals that do not cross such points. The absolute value in \(\ln|g(x)|\) allows the logarithm formula to work whether \(g(x)\) is positive or negative on the interval.
In the second formula, the exponent \(1-n\) is negative. For example, a denominator to the third power produces a result involving the reciprocal square of that denominator's inner expression. The constant \(C\) is included because any two antiderivatives on an interval differ by a constant.
Why the Formulas Work
The justification is the substitution rule, which reverses the chain rule. In the first-power case, differentiate the proposed logarithm. The derivative of \(\ln|g(x)|\), wherever \(g(x)\ne0\), is \(g'(x)/g(x)\). Multiplying by \(k\) gives exactly the integrand.
For a higher power, differentiate the proposed expression using the power rule and chain rule. The exponent \(1-n\) reduces by one, while the coefficient \(1-n\) from the power rule cancels the denominator \(1-n\) in the proposed antiderivative. What remains is the original integrand.
This also explains the distinction between the cases. The power rule for an antiderivative of \(u^p\) requires \(p\ne-1\). At \(n=1\), the substituted power is \(u^{-1}\), the exceptional exponent, and its antiderivative is \(\ln|u|\) instead.
Worked Examples
Worked Example: A First-Power Composite Denominator
Evaluate \(\int \frac{6x-2}{3x^2-2x+5}\,dx\).
Let \(g(x)=3x^2-2x+5\). Its derivative is \(g'(x)=6x-2\), exactly the numerator. Set \(u=3x^2-2x+5\), so \(du=(6x-2)\,dx\). Therefore, \(\int \frac{6x-2}{3x^2-2x+5}\,dx=\int \frac{1}{u}\,du=\ln|u|+C=\ln|3x^2-2x+5|+C\).
To check, differentiate the answer: \(\frac{d}{dx}\ln|3x^2-2x+5|=\frac{6x-2}{3x^2-2x+5}\). The derivative agrees with the integrand. In fact, the quadratic is positive for all real \(x\), since its discriminant is \((-2)^2-4(3)(5)=4-60=-56<0\) and its leading coefficient is positive. The absolute value is still appropriate in the general formula.
Worked Example: A Cubed Denominator
Evaluate \(\int \frac{4x}{(2x^2+7)^3}\,dx\).
Choose the inner expression \(g(x)=2x^2+7\). Its derivative is \(g'(x)=4x\), which matches the numerator. Set \(u=2x^2+7\) and \(du=4x\,dx\). Then \(\int \frac{4x}{(2x^2+7)^3}\,dx=\int u^{-3}\,du=\frac{u^{-2}}{-2}+C=-\frac{1}{2u^2}+C\). Substituting back gives \(-\frac{1}{2(2x^2+7)^2}+C\).
Differentiate to check: \(\frac{d}{dx}\left[-\frac12(2x^2+7)^{-2}\right]=-\frac12(-2)(2x^2+7)^{-3}(4x)=\frac{4x}{(2x^2+7)^3}\). This is the original integrand.
Worked Example: A Constant Multiple in the Numerator
Evaluate \(\int \frac{10x^4}{(x^5+3)^4}\,dx\).
Take \(g(x)=x^5+3\), so \(g'(x)=5x^4\). The numerator is \(10x^4=2g'(x)\), so the constant factor is \(k=2\). Set \(u=x^5+3\), and then \(du=5x^4\,dx\). Since \(10x^4\,dx=2\,du\), the integral becomes \(2\int u^{-4}\,du\).
Integrate the power: \(2\int u^{-4}\,du=2\left(\frac{u^{-3}}{-3}\right)+C=-\frac{2}{3}u^{-3}+C\). Substitution back gives \(-\frac{2}{3(x^5+3)^3}+C\).
To verify, differentiate: \(\frac{d}{dx}\left[-\frac23(x^5+3)^{-3}\right]=-\frac23(-3)(x^5+3)^{-4}(5x^4)=\frac{10x^4}{(x^5+3)^4}\). The original integrand is recovered.
A Reliable Procedure
For an integral that appears to have this form, use the following sequence. The crucial check is that the numerator, including its constant factor, really equals the derivative of the denominator's inner expression.
Write the denominator as a power of a function \(g(x)\), and find \(g'(x)\).
Check whether it is exactly \(k g'(x)\) for a constant \(k\). Do not assume a match just because the denominator is composite.
Set \(u=g(x)\) and replace \(g'(x)\,dx\) by \(du\), retaining any constant factor.
Use \(\ln|u|\) when the denominator has power one, and the power rule when the power is greater than one.
Express the result using \(x\), include \(+C\), and differentiate to confirm the integrand.
Common Pitfall: Missing the Factor or the Power
A common error is to recognize the right denominator but drop the constant relating its derivative to the numerator. For example, in \(\int \frac{10x^4}{(x^5+3)^4}\,dx\), the derivative of \(x^5+3\) is \(5x^4\), not \(10x^4\). The substitution must preserve the factor of \(2\): \(10x^4\,dx=2\,du\).
Another error is to use the logarithm formula for every denominator power. That formula applies to \(g'(x)/g(x)\), not to \(g'(x)/[g(x)]^4\). For the latter, the substituted integrand has exponent \(-4\), so the power rule gives an exponent of \(-3\) and a coefficient of \(-1/3\). A derivative check quickly exposes either error.
For instance, incorrectly treating the preceding example as a logarithm would give \(2\ln|x^5+3|\). Its derivative is \(2(5x^4)/(x^5+3)=10x^4/(x^5+3)\), which has only a first power in the denominator. It does not equal the original integrand, whose denominator has the fourth power.
When This Direct Pattern Does Not Apply
The method depends on a specific match: the numerator must be a constant multiple of the derivative of the denominator's inner expression. If that match is absent, replacing the inner expression by \(u\) does not turn the entire integrand into a power of \(u\) times \(du\). Do not force the substitution or discard leftover factors. The direct formulas above apply only when the derivative match is complete.
Keeping that limitation in mind makes the method more useful, not less. It distinguishes a genuine reverse-chain-rule pattern from an integrand that merely has a complicated denominator. When the match is present, the calculation is short; when it is absent, another approach is needed.
Why the Pattern Matters
Composite denominators occur frequently, and their powers can make an integral look more complicated than it is. Recognizing the derivative match reduces the expression to one of two basic forms: \(1/u\), whose antiderivative is logarithmic, or \(u^{-n}\) for \(n>1\), whose antiderivative follows the power rule.
The method also reinforces a central calculus habit: check the result in the direction opposite to the calculation. Substitution proposes an antiderivative; differentiating that proposal verifies it. In this pattern, the chain rule makes the check direct and confirms both the exponent and any constant factor.
Check Your Understanding
For each question, identify the denominator's inner expression and determine whether the numerator matches its derivative up to a constant.
- Evaluate \(\int \frac{3x^2}{x^3+4}\,dx\), and verify by differentiation.
- Evaluate \(\int \frac{6x}{(3x^2+1)^2}\,dx\). Which power rule exponent appears after substitution?
- For \(\int \frac{8x^3}{(2x^4-5)^3}\,dx\), find \(g(x)\), \(g'(x)\), and the constant \(k\).
- Why is \(\ln|g(x)|\) used for a first-power denominator rather than the ordinary power formula with exponent \(-1\)?
- Does the direct pattern apply to \(\int \frac{x}{(x^2+1)^2}\,dx\)? Explain by comparing the numerator with the derivative of the inner expression.
