A Reliable Route from Rational Function to Integral
When the degree of a rational function’s numerator is at least the degree of its denominator, the fraction is called improper. Partial fractions apply directly to a proper rational function, whose numerator has smaller degree than its denominator. For an improper rational integral with repeated linear factors, the route is therefore: divide first, decompose the proper remainder, and integrate the resulting terms.
The repeated factors matter because a complete decomposition must include a term for every power of each factor. For instance, a denominator \((x-a)^3\) requires terms with denominators \(x-a\), \((x-a)^2\), and \((x-a)^3\). Once the terms are found, the antiderivatives come from the logarithm rule for the first power and the power rule for higher powers.
The Integration Rules We Need
For \(x\neq a\), integrating a simple linear-factor term gives a logarithm. For an integer \(k\geq2\), integrating a higher power gives a negative power. These formulas follow by setting \(u=x-a\), so \(du=dx\), and applying the logarithm and power rules:
To verify the second formula, differentiate its right-hand side:
For a definite integral, the integrand must be defined throughout the interval for the usual evaluation by an antiderivative to apply. In the examples below, each interval avoids the zeros of the denominator.
Worked Example: An Indefinite Integral with a Repeated Factor
Worked Example: Divide and Integrate a Cubic Rational Function
Find an antiderivative of \(\dfrac{x^3+1}{(x-1)^2}\). The numerator has degree \(3\), while the denominator has degree \(2\), so we begin with polynomial division. Since \((x-1)^2=x^2-2x+1\), the first quotient term is \(x\):
The next quotient term is \(2\). Subtracting \(2(x^2-2x+1)\) from the current remainder gives:
Thus the quotient is \(x+2\), and the remainder is \(3x-1\):
The proper remainder has the form \(A/(x-1)+B/(x-1)^2\). Multiplying through by \((x-1)^2\) gives a polynomial identity:
Matching coefficients gives \(A=3\) and \(-A+B=-1\), so \(B=2\). Therefore,
Integrate each term. The polynomial terms give \(x^2/2+2x\); the simple factor gives \(3\ln|x-1|\); and the repeated-factor term gives \(-2/(x-1)\):
Check by differentiating. The derivative of \(-2/(x-1)\) is \(2/(x-1)^2\), so the derivative of the proposed antiderivative is \(x+2+3/(x-1)+2/(x-1)^2\), exactly the decomposition above. Recombining its proper terms gives \((3x-1)/(x-1)^2\), and adding the quotient \(x+2\) recovers the original integrand.
Why Division Must Come First
Polynomial division separates the part that is already polynomial from the proper rational part. This is what makes partial fractions applicable: the remainder’s degree is strictly less than the denominator’s degree. Integration then follows term by term, using the sum rule for integrals.
The algebraic identity at the center of the method is:
If \(Q(x)\) is a product of repeated linear factors, the proper fraction \(R(x)/Q(x)\) can be decomposed into terms for the successive powers of each factor. The coefficient identity is checked by multiplying through by \(Q(x)\); equality of the resulting polynomials guarantees equality of the rational expressions wherever \(Q(x)\neq0\). The integral is consequently the sum of the polynomial integral and the integrals of the partial fractions.
Write the improper rational function as a polynomial plus a proper rational function.
Include a term for every required power of each repeated linear factor, then solve the coefficient identity.
Integrate the polynomial, logarithmic terms, and higher-power terms separately.
For a definite integral, confirm the interval contains no pole, substitute the endpoints, and check the decomposition or antiderivative.
Worked Example: A Definite Integral with a Cubed Factor
Worked Example: Evaluate on an Interval Without a Pole
Evaluate \(\displaystyle\int_0^1 \frac{x^4}{(x+1)^3}\,dx\). The numerator degree \(4\) is greater than the denominator degree \(3\). The only denominator zero is \(x=-1\), which is outside \([0,1]\), so the integrand has no pole on the interval.
Divide \(x^4\) by \((x+1)^3=x^3+3x^2+3x+1\). The first quotient term is \(x\), leaving \(-3x^3-3x^2-x\). The next quotient term is \(-3\), leaving:
Thus the quotient is \(x-3\), and the remainder is \(6x^2+8x+3\). Write the proper fraction as \(A/(x+1)+B/(x+1)^2+C/(x+1)^3\). Multiplying by \((x+1)^3\) gives:
Expanding the right-hand side gives \(Ax^2+(2A+B)x+(A+B+C)\). Therefore \(A=6\), \(2A+B=8\), and \(A+B+C=3\). These equations give \(B=-4\) and \(C=1\), so:
An antiderivative is:
Here \(x+1>0\) on the interval, so \(\ln(x+1)=\ln|x+1|\). Evaluating at the endpoints gives:
Subtracting, the exact value is:
For a check, differentiating \(F\) gives \(x-3+6/(x+1)-4/(x+1)^2+1/(x+1)^3\). Combining these terms over \((x+1)^3\) gives numerator \(x^4\), as in the original integrand.
Worked Example: A Definite Integral with a Squared Factor
Worked Example: Evaluate a Rational Function on a Pole-Free Interval
Evaluate \(\displaystyle\int_0^1 \frac{x^2+1}{(x-2)^2}\,dx\). The denominator is nonzero on \([0,1]\), since its only zero is \(x=2\).
Since \((x-2)^2=x^2-4x+4\), subtracting this denominator polynomial from the numerator gives the remainder \(4x-3\). Thus:
Write the proper fraction as \(A/(x-2)+B/(x-2)^2\). Clearing denominators gives \(4x-3=A(x-2)+B=Ax-2A+B\). Hence \(A=4\), and \(-2A+B=-3\) gives \(B=5\). The full decomposition is:
Integrating gives the antiderivative \(F(x)=x+4\ln|x-2|-5/(x-2)\). At \(x=1\), \(F(1)=1+0+5=6\). At \(x=0\), \(F(0)=0+4\ln2+5/2\). Therefore:
Check by differentiation: \(F'(x)=1+4/(x-2)+5/(x-2)^2\). Putting these terms over \((x-2)^2\) gives numerator \((x-2)^2+4(x-2)+5=x^2+1\), which verifies the original integrand.
A Common Mistake: Treating Every Power Like a Logarithm
A frequent error is to integrate a repeated-factor term as if it had denominator \((x-a)\). For example, the following calculation is wrong:
Differentiating the incorrect result gives \(2/(x-1)\), not \(2/(x-1)^2\). The correct calculation uses the power rule:
Differentiation confirms the correction: \(\dfrac{d}{dx}\bigl(-2/(x-1)\bigr)=2/(x-1)^2\). The logarithm belongs to the first power only; every higher integer power uses the power rule.
Putting the Method to Work
In an improper rational integral with repeated linear factors, keep the algebra and calculus stages distinct. First make the rational function proper by division. Then solve the partial-fraction identity for the remainder, retaining all successive powers of each repeated factor. Finally, integrate and, for a definite integral, check that no denominator zero lies in the interval before evaluating the antiderivative at the endpoints.
Check Your Understanding
Use the complete sequence of division, decomposition, integration, and checking where appropriate.
- Why must polynomial division precede partial fractions when the numerator degree is at least the denominator degree?
- What denominator terms are needed for a proper fraction whose denominator contains \((x-3)^4\)?
- Find \(\displaystyle\int \frac{1}{(x+2)^3}\,dx\), and check your answer by differentiation.
- Before evaluating \(\displaystyle\int_a^b P(x)/Q(x)\,dx\) by an antiderivative, what must be checked about the zeros of \(Q\) on \([a,b]\)?
- In a coefficient identity for \(A/(x-1)+B/(x-1)^2\), why is it valid to match coefficients after multiplying through by \((x-1)^2\)?
