Why the Exponent Matters
A repeated linear factor can appear in a rational function as a term such as \(A/(ax+b)^k\), where the denominator is raised to a positive integer power. Integrating these terms is a routine step after a rational function has been decomposed into simpler pieces. The exponent determines which rule applies: a first power gives a logarithm, while higher powers give a power-rule antiderivative.
The coefficient \(a\) inside the linear expression matters too. It changes the answer by a factor of \(1/a\). This is the same chain-rule issue that arises whenever an antiderivative contains a function of a linear expression. Leaving out that factor produces an answer whose derivative is too large or too small.
The Formulas and Their Conditions
Assume \(A\) and \(a\) are constants, \(a\ne 0\), \(b\) is a constant, and \(k\) is a positive integer. The formulas below apply on any interval where \(ax+b\ne 0\). The absolute value in the logarithm allows the formula to apply on either side of the point where the linear expression is zero.
Here \(C\) is an arbitrary constant. Since \(1-k\ne 0\) when \(k>1\), the second formula has a nonzero denominator. It must not be used with \(k=1\): that would require division by zero. The first-power case is genuinely different and yields the logarithm.
If \(A=0\), the integrand is zero and its antiderivative is a constant. If \(a=0\), then the denominator is constant rather than a linear factor, so these formulas do not apply; when \(b\ne 0\), the integral is simply \(Ax/b^k+C\).
Why the Formulas Work
The formulas can be checked by differentiation. That check also explains why the factor \(1/a\) is necessary. For \(k>1\), differentiate the proposed nonconstant part using the power rule and the chain rule:
The factors \(1-k\) and \(a\) from differentiation cancel the corresponding factors in the denominator. This proves that the displayed expression is an antiderivative on any interval where it is defined.
For \(k=1\), the power-rule form cannot be used. Instead, the derivative of the logarithmic expression gives
Thus both formulas follow from differentiation, but they serve different exponents. The logarithm is not an optional rewrite of the higher-power formula; it is required for the case \(k=1\).
Worked Examples
Worked Example: Integrating a Sum of Repeated Linear Terms
Find an antiderivative of
Integrate each term separately. For the first term, \(A=6\), \(a=2\), and \(k=3\). Its coefficient in the power-rule formula is \(6/[2(1-3)]=-3/2\). For the second, \(A=-4\), \(a=3\), and \(k=2\), giving coefficient \(-4/[3(1-2)]=4/3\). The last term has exponent one, so use the logarithmic formula with coefficient \(5/5=1\). Therefore, an antiderivative is
To check, differentiate the first term: \((-3/2)(-2)(2)(2x+1)^{-3}=6/(2x+1)^3\). Differentiating the second gives \((4/3)(-1)(3)(3x-2)^{-2}=-4/(3x-2)^2\). Differentiating the logarithm gives \(5/(5x+4)\). Their sum is exactly the original integrand.
Worked Example: A Decreasing Linear Expression
Find an antiderivative of
The linear expression is \(4-3x\), so \(a=-3\), not \(3\). Here \(A=-7\) and \(k=4\). The coefficient in the formula is
Thus an antiderivative is
Differentiate to check: the derivative of \((4-3x)^{-3}\) is \((-3)(4-3x)^{-4}(-3)=9(4-3x)^{-4}\). Multiplying by \(-7/9\) gives \(-7/(4-3x)^4\), as required.
Worked Example: Combining a First Power and Higher Powers
Find an antiderivative of
For the first term, \(A=2\), \(a=1\), and \(k=2\), so the coefficient is \(2/[1(1-2)]=-2\). For the second, \(A=3\), \(a=2\), and \(k=3\), so the coefficient is \(3/[2(1-3)]=-3/4\). The last term has \(A=-6\), \(a=4\), and \(k=1\); its logarithmic coefficient is \(-6/4=-3/2\). The result is
Differentiate each part. The first derivative is \(2(x+2)^{-2}\). The second is \((-3/4)(-2)(2)(2x-1)^{-3}=3(2x-1)^{-3}\). The last derivative is \((-3/2)(4)/(4x+5)=-6/(4x+5)\). Adding these derivatives recovers every term of the integrand, so the answer checks.
Using the Rule Term by Term
Once a rational expression has been written as a sum of terms involving repeated linear factors, integrate each term separately and then add the results. This uses the linearity of integration: the antiderivative of a finite sum is the sum of antiderivatives, with one overall arbitrary constant. A useful way to organize the work is to identify \(A\), \(a\), and \(k\) for each term before choosing a formula.
Write the denominator as \((ax+b)^k\) and record the coefficient \(a\) of \(x\), including its sign.
If \(k=1\), use \((A/a)\ln|ax+b|\). If \(k>1\), use \(A(ax+b)^{1-k}/[a(1-k)]\).
Apply the selected formula to each term, then differentiate the result to verify both the exponent and the chain-rule factor.
The formulas describe antiderivatives on intervals that do not cross a zero of a denominator. For example, a term with denominator \(2x-1\) is undefined at \(x=1/2\). Its logarithmic antiderivative is defined separately on intervals to either side of that point. This domain condition is not a change to the formula; it records where the integrand and its antiderivative are defined.
A Common Error: Forgetting the Inner Factor
A frequent mistake is to use the power rule on \((2x+3)^{-3}\) but overlook the derivative \(2\) of the inner expression. For instance, a tempting but incorrect calculation is
Differentiating that proposed answer gives \((-5/2)(-2)(2)(2x+3)^{-3}=10/(2x+3)^3\), which is twice the desired integrand. The corrected coefficient includes division by \(a=2\): \(5/[2(1-3)]=-5/4\). Hence
Its derivative is \((-5/4)(-2)(2)(2x+3)^{-3}=5/(2x+3)^3\), confirming the correction. Checking by differentiation is especially effective here because a missing inner factor appears directly as a wrong multiplicative coefficient.
Why This Distinction Is Useful
Repeated linear-factor terms arise in the integration of rational functions, and their exponents can differ within the same expression. Treating every term with a logarithm would give incorrect results for powers greater than one; applying the power rule indiscriminately would fail at exponent one. Classifying the exponent first prevents both errors.
For a quick decision, compare the exponent and the resulting form:
| Denominator term | Antiderivative form |
|---|---|
| \(A/(ax+b)\) | \((A/a)\ln|ax+b|+C\) |
| \(A/(ax+b)^k\), \(k>1\) | \(A(ax+b)^{1-k}/[a(1-k)]+C\) |
In either row, the safest final step is the same: differentiate the proposed antiderivative. The result should reproduce the original numerator, denominator power, and sign exactly.
Check Your Understanding
For each question, state the relevant exponent and account for the coefficient of \(x\) inside the linear expression. Verify your results by differentiation.
- Find an antiderivative of \(4/(3x+1)^2\).
- Find an antiderivative of \(-2/(5-2x)^3\).
- Why does the power-rule formula for \(k>1\) not apply when \(k=1\)?
- Find an antiderivative of \(3/(2x+7)\), including the correct logarithm and coefficient.
- On what points is an antiderivative for \(1/(4x-1)\) not defined by its logarithmic formula?
