From Rational Expressions to Definite Integrals
A rational integrand can often be rewritten as a polynomial plus simpler fractions. That rewrite makes antiderivatives manageable, but evaluating a definite integral requires one more check: the rational function must be defined throughout the interval, including its endpoints. A denominator zero may cause a pole, or it may disappear after a common factor is canceled. The reduced fraction determines which situation applies.
Once the integrand is continuous on the closed interval, its decomposition can be integrated term by term. The resulting antiderivative is evaluated at the two endpoints, and differentiating it provides a direct check of the calculation.
A zero of the original denominator is not automatically a pole. If a common factor cancels, the expression may instead have a removable singularity at that point. If, after cancellation, the denominator still vanishes while the numerator does not, the reduced rational function has a pole there. Check the reduced fraction before deciding whether the ordinary definite-integral procedure applies.
Why Division and Decomposition Can Be Used
For a rational function whose numerator has degree at least as large as its denominator, polynomial division writes it as a polynomial plus a proper rational function. Partial fractions can then express the proper part as a sum of simpler terms. When the reduced denominator has no zeros on \([a,b]\), these equalities hold throughout the interval, and the component antiderivatives can be combined.
The justification is the linearity of differentiation and the Fundamental Theorem of Calculus. If \(F_i'(x)\) equals the corresponding term in the decomposition, then the sum \(F=F_S+F_1+\cdots+F_n\) satisfies \(F'(x)=R(x)\) throughout the interval. Since \(R\) is continuous on \([a,b]\), the Fundamental Theorem gives \(\int_a^b R(x)\,dx=F(b)-F(a)\). The continuity condition matters: a decomposition does not make an integral across a pole an ordinary definite integral.
Worked Example: Division Followed by a Logarithm
Worked Example: Divide and integrate on an interval without a pole
Evaluate \(\int_0^1 \frac{x^2+1}{x-2}\,dx\). The denominator \(x-2\) is nonzero on \([0,1]\), since \(x-2\le -1\) there. There is no common factor to cancel.
Divide the numerator by the denominator. The quotient is \(x+2\), because
Thus the integrand is
An antiderivative is \(F(x)=\frac{x^2}{2}+2x+5\ln|x-2|\). Evaluate each endpoint:
Therefore,
To check, differentiate \(F\):
The decomposition and antiderivative are correct. The result is negative, as expected, because the numerator is positive and the denominator is negative throughout \([0,1]\).
Worked Example: A Quotient and Two Linear Factors
Worked Example: Divide, decompose, and evaluate at the endpoints
Evaluate \(\int_0^1 \frac{x^3+1}{x^2+5x+6}\,dx\). Factor the denominator as \((x+2)(x+3)\). Its zeros, \(-2\) and \(-3\), lie outside \([0,1]\), so the denominator is nonzero throughout the interval.
First perform polynomial division. The quotient \(x-5\) gives
Subtracting this product from the numerator leaves \(19x+31\), so
Write the proper fraction as \(A/(x+2)+B/(x+3)\). Matching coefficients in \(19x+31=A(x+3)+B(x+2)\) gives \(A+B=19\) and \(3A+2B=31\). Subtracting twice the first equation from the second gives \(A=-7\), and then \(B=26\). Therefore,
An antiderivative is \(F(x)=\frac{x^2}{2}-5x-7\ln|x+2|+26\ln|x+3|\). Since both linear factors are positive on the interval, the logarithms can be written without absolute-value bars when evaluating there. The endpoint values are
Subtract and combine logarithms:
For the check, differentiate the proposed antiderivative and recombine:
The derivative recovers the original integrand, and the endpoint subtraction gives the definite integral.
Worked Example: A Quadratic Denominator
Worked Example: Separate a logarithmic part from an arctangent part
Evaluate \(\int_0^1 \frac{2x+3}{x^2+2x+5}\,dx\). Complete the square in the denominator:
for every real \(x\). In particular, the denominator has no zeros on \([0,1]\). The numerator can be separated as \(2x+3=(2x+2)+1\), giving
The first term has antiderivative \(\ln(x^2+2x+5)\), since the derivative of the denominator is \(2x+2\). The second has antiderivative \(\frac12\arctan\!\left(\frac{x+1}{2}\right)\), as follows by the chain rule. Thus
At the endpoints, \(F(1)=\ln 8+\frac12\arctan(1)=\ln 8+\frac{\pi}{8}\), while \(F(0)=\ln 5+\frac12\arctan(1/2)\). Hence
Differentiate to verify the antiderivative:
The equality follows because \((x+1)^2+4=x^2+2x+5\). This confirms both the split and the endpoint evaluation.
Check the Reduced Denominator, Not Just the Original One
Here is a common mistake: “The denominator is zero at \(x=0\), so the integrand has a pole there.” That conclusion is not justified without checking for a common factor. For instance, for \(x\ne0\),
The original expression is undefined at zero, but canceling the common factor leaves the constant function \(1\), whose continuous extension has value \(1\) at zero. The point is a removable singularity, not a pole of the reduced rational function. If the integrand is defined at zero by that extension, its integral on \([-1,1]\) is \(\int_{-1}^1 1\,dx=2\). If no value is assigned there, the original expression is not defined at every point of the interval; one must state that the integral is being taken using the continuous extension (or as an improper integral).
In contrast, if a factor remains in the reduced denominator and vanishes within the interval, the reduced rational function has a pole there. Do not apply the ordinary endpoint-antiderivative evaluation across that point as though the function were continuous. The location and nature of the singularity determine what kind of integral is appropriate.
A separate endpoint pitfall is to substitute into \(\ln|g(x)|\) without checking whether \(g\) vanishes at an endpoint. In the examples above, the denominators were verified to be nonzero on the entire closed interval, so every logarithm was evaluated at a nonzero argument. Taking absolute values handles negative arguments; it does not make \(\ln 0\) defined.
A Reliable Evaluation Sequence
Cancel common factors, then check whether the reduced denominator is nonzero at every point of the closed interval.
If the numerator degree is at least the denominator degree, use polynomial division to separate a polynomial quotient from a proper remainder.
Express the proper rational part as simpler terms and integrate each term using its applicable antiderivative.
Subtract the value at the lower endpoint from the value at the upper endpoint, then differentiate and recombine to check the antiderivative.
The key discipline is to establish that the reduced function is continuous on the interval before using endpoint evaluation. Division and partial fractions provide an algebraic route to an antiderivative; the denominator check ensures that the route is valid across the whole interval.
Check Your Understanding
Answer each question by identifying the relevant condition or showing the essential calculation.
- For \(\frac{x^2+1}{x-2}\) on \([0,1]\), why is endpoint evaluation of an antiderivative valid?
- Divide \(x^2+3x+1\) by \(x+1\), and state the quotient and remainder.
- What must be checked before concluding that a zero of the original denominator is a pole?
- Differentiate \(\ln(x^2+1)\) and state the numerator that results over the denominator \(x^2+1\).
- Why is it not valid to use ordinary endpoint-antiderivative evaluation across a pole inside the interval?
