Why Logarithms Appear
A simple linear partial-fraction term has a constant numerator and a linear denominator. Its integral is governed by the logarithm: differentiating a logarithm of the denominator recovers the reciprocal linear factor. For a definite integral, the result depends on the values at both endpoints, so it is essential to keep track of where the denominator is zero and to use absolute values correctly.
This tutorial considers ordinary definite integrals over finite closed intervals. The denominator of each nonzero term must stay nonzero throughout the interval. If a denominator vanishes there, the integral is not an ordinary finite-interval integral of a continuous function, and the formula developed here does not apply.
The Antiderivative of a Simple Linear Term
For \(C\ne 0\), and for \(x\ne a\), an antiderivative is \(C\ln|x-a|\). The absolute value matters because \(x-a\) may be positive or negative. The real logarithm is defined only for positive inputs, but \(|x-a|>0\) whenever \(x\ne a\).
Here \(K\) is an arbitrary constant. The formula also gives zero for \(C=0\), wherever the original expression is defined. In a definite integral, the constant \(K\) cancels when the endpoint values are subtracted.
Why the Logarithmic Formula Works
The key derivative can be checked on both sides of \(a\). If \(x>a\), then \(|x-a|=x-a\), so the derivative of \(\ln|x-a|\) is \(1/(x-a)\). If \(x<a\), then \(|x-a|=a-x\), and differentiating \(\ln(a-x)\) gives \(-1/(a-x)=1/(x-a)\). Thus the same derivative formula holds on either interval that does not contain \(a\).
On an interval that does not contain \(a\), the function \(C/(x-a)\) is continuous.
For every \(x\ne a\), the chain rule and the absolute-value calculation above give \( \frac{d}{dx}\bigl(C\ln|x-a|\bigr)=C/(x-a) \).
On a finite interval \([u,v]\) that avoids \(a\), the integrand is continuous and the logarithm is an antiderivative there. Therefore its endpoint difference equals the definite integral.
The condition excludes a pole at either endpoint as well as a pole in the interior. It guarantees that every logarithm in the endpoint formula has a positive, nonzero argument and that the integrand is continuous on the whole closed interval.
Evaluating One Term
Worked Example: A Positive Denominator Throughout the Interval
Evaluate \( \int_0^2 \frac{3}{x+1}\,dx \). Here \(x+1=x-(-1)\), so the pole is at \(x=-1\), outside \([0,2]\). An antiderivative is \(F(x)=3\ln|x+1|\).
Substitute the upper endpoint: \(F(2)=3\ln|3|=3\ln 3\). Substitute the lower endpoint: \(F(0)=3\ln|1|=3\ln 1=0\). Subtract lower from upper:
Check by differentiating the proposed antiderivative: \(F'(x)=3\cdot\frac{1}{x+1}\cdot 1=3/(x+1)\), the original integrand. The pole is outside the interval, so the endpoint evaluation is valid.
Absolute Values and Intervals to the Left of a Pole
An interval need not lie to the right of a pole. If it lies entirely to the left, \(x-a\) is negative throughout, but the absolute value still produces a positive logarithm input. The derivative remains \(1/(x-a)\); changing the sign of the denominator does not change the antiderivative rule.
Worked Example: An Interval Entirely to the Left of the Pole
Evaluate \( \int_{-3}^{-1} \frac{2}{x-1}\,dx \). The pole is \(x=1\), which is not in \([-3,-1]\). An antiderivative is \(F(x)=2\ln|x-1|\).
At the upper endpoint, \(F(-1)=2\ln|-2|=2\ln 2\). At the lower endpoint, \(F(-3)=2\ln|-4|=2\ln 4\). Therefore,
For a check, differentiating \(2\ln|x-1|\) gives \(2/(x-1)\) for every \(x\ne 1\). Also, the integrand is negative throughout this interval, consistent with the negative value of the integral.
Integrating a Sum of Simple Linear Terms
Integration is linear: the integral of a finite sum is the sum of the integrals, and constant coefficients can be factored out. Thus a sum of simple linear terms is handled by finding one logarithmic antiderivative for each term, adding them, and evaluating the resulting expression at the endpoints. Combine duplicate terms first if necessary; the formula applies to the resulting terms.
Worked Example: Two Terms with Different Poles
Evaluate \( \int_0^1\left(\frac{2}{x-2}-\frac{1}{x+1}\right)\,dx \). The poles are \(2\) and \(-1\), both outside \([0,1]\). Integrating each term gives the antiderivative \(F(x)=2\ln|x-2|-\ln|x+1|\).
At \(x=1\), the two logarithm inputs are \(|1-2|=1\) and \(|1+1|=2\), so \(F(1)=2\ln 1-\ln 2=-\ln 2\). At \(x=0\), the inputs are \(|0-2|=2\) and \(|0+1|=1\), so \(F(0)=2\ln 2-\ln 1=2\ln 2\). Hence,
Differentiate to check: \(F'(x)=2/(x-2)-1/(x+1)\), exactly the integrand. The signs also agree: on \([0,1]\), \(2/(x-2)<0\) and \(-1/(x+1)<0\), so a negative integral is expected.
A Common Error: Dropping the Absolute Value
A tempting but invalid step is to use \(\ln(x-a)\) when \(x-a\) is negative. For example, on \([0,1]\), the expression \(x-2\) is negative. The calculation \( \ln(1-2)-\ln(0-2)=\ln(-1)-\ln(-2) \) is not a valid calculation with real logarithms: neither logarithm is defined.
The correction is to retain the absolute values in the antiderivative and at both endpoints. For this same integrand,
Indeed, \( \frac{d}{dx}\ln|x-2|=1/(x-2) \) on this interval. The absolute value makes the logarithm real; it does not remove the negative sign of the original integrand.
When the Formula Does Not Apply
If a nonzero term \(C/(x-a)\) has \(a\in[u,v]\), its denominator vanishes somewhere on the interval. The function is then not continuous on the entire closed interval, so the ordinary finite-interval formula from the theorem cannot be used. In particular, it is not valid simply to substitute the two endpoints into a logarithmic expression and ignore a pole between them. Integrals involving such singularities require separate treatment and are outside the scope of this tutorial.
Always check the poles of the actual terms being integrated. A term with \(C=0\) contributes zero and is omitted when checking for poles; a term with \(C\ne0\) has a pole at its denominator’s zero. This distinction avoids incorrectly treating a zero term as having a pole.
Check Your Understanding
Use the logarithmic antiderivative and check that each interval avoids the poles of all nonzero terms.
- Find an antiderivative of \( \frac{5}{x+3} \) on an interval that does not contain \(x=-3\).
- Evaluate \( \int_0^2 \frac{1}{x-4}\,dx \) in exact form.
- Evaluate \( \int_{-2}^{-1} \frac{3}{x+1}\,dx \), or explain why the ordinary finite-interval formula does not apply.
- For what values of \(x\) is \( \frac{d}{dx}\ln|x-5|=\frac{1}{x-5} \) valid?
- If a partial-fraction term is \(0/(x-2)\), should \(x=2\) be counted as a pole of that term? Explain.
