Why Rewrite a Rational Expression?
A rational expression is a quotient of polynomials. Some such expressions can be integrated directly after a small algebraic rewrite: separate the numerator’s terms, express negative powers explicitly, or divide a linear numerator by a linear denominator. The result is a sum of familiar powers and reciprocal terms.
This matters because the denominator’s appearance can hide a simple structure. For example, a fraction with \(x^2\) in the denominator may contain a constant term, a reciprocal term, and a power term. Once those pieces are visible, the power rule for antiderivatives and the logarithmic antiderivative apply term by term.
Basic Forms and Their Conditions
For a real exponent \(r\ne -1\), the power rule gives an antiderivative of \(x^r\) on intervals where the expression is defined. The reciprocal case \(r=-1\) has a different form: its antiderivative is \(\ln|x|\), valid on intervals not containing zero.
Here \(C\) is an arbitrary constant. The absolute value in \(\ln|x|\) allows the formula to apply on both intervals \(x>0\) and \(x<0\). The function \(\ln|x|\) is not defined at \(x=0\), just as \(1/x\) is not defined there.
A linear denominator \(ax+b\), where \(a\ne0\), has a closely related reciprocal form. The factor \(1/a\) is needed because the derivative of \(ax+b\) is \(a\).
For any of these rules, the integrand must be defined on the interval under consideration. If the original rational expression is undefined at a point, an antiderivative is considered separately on each interval of its domain; one constant does not have to work across a point where the integrand is undefined.
Rewrite Powers of \(x\) in the Denominator
When a polynomial is divided by \(x^n\), divide each numerator term by the denominator. Equivalently, use \(x^k/x^n=x^{k-n}\), provided \(x\ne0\). This converts the fraction into a sum of powers, including negative powers.
For instance, dividing each term of \(3x^2+2x-5\) by \(x^2\) gives \(3+2x^{-1}-5x^{-2}\). The first two powers integrate by the power and logarithmic rules; the final term uses the power rule with exponent \(-2\).
Worked Example: A Polynomial over \(x^2\)
Worked Example: A Polynomial over \(x^2\)
Find an antiderivative of \(\displaystyle \frac{3x^2+2x-5}{x^2}\), and check it.
The denominator is zero at \(x=0\), so work on an interval that does not contain zero. Divide each numerator term by \(x^2\):
Integrate each term. Write \(x^{-2}\) as a power, and use the logarithmic rule for \(1/x\):
Check by differentiation on either interval \(x<0\) or \(x>0\):
The derivative equals the original integrand wherever that integrand is defined, so the antiderivative checks.
Rewrite a Linear Numerator over a Linear Denominator
If both numerator and denominator are linear, express the numerator as a multiple of the denominator plus a constant. For \((mx+n)/(ax+b)\), choose a number \(q\) so that \(mx\) matches the \(x\)-term in \(q(ax+b)\). Then the leftover is a constant. Dividing by \(ax+b\) produces a constant term plus a reciprocal term.
This is ordinary algebraic division, not a new integration rule. The resulting reciprocal term is handled by the logarithmic formula. The denominator must not be zero; if \(a\ne0\), its zero is \(x=-b/a\).
Worked Example: A Linear Denominator
Worked Example: A Linear Denominator
Find an antiderivative of \(\displaystyle \frac{4x+1}{2x-3}\), and check it.
Choose \(2\) as the quotient coefficient, since \(2(2x)=4x\). Subtract \(2(2x-3)\) from the numerator to find the remainder:
Thus the numerator is \(2(2x-3)+7\), and the fraction becomes
The denominator is zero at \(x=3/2\). On either interval that avoids this point, integrate the constant and the linear reciprocal. Since the coefficient of \(x\) in \(2x-3\) is \(2\), the logarithmic antiderivative includes a factor \(1/2\):
Differentiate to verify both terms:
The derivative reproduces the original expression, and the antiderivative is valid on intervals that do not cross \(x=3/2\).
Why the Rewriting Method Works
The method rests on two facts. First, the algebraic rewrite gives an identity of functions wherever the original denominator is nonzero. Second, integration is linear: the antiderivative of a sum is the sum of antiderivatives, and constant factors can be taken outside the integral. These statements explain why each separated term can be integrated using its corresponding rule.
The logarithmic formula for a linear denominator can also be justified directly. On an interval where \(ax+b\ne0\), apply the chain rule to \(\ln|ax+b|\): its derivative is \(a/(ax+b)\). Multiplication by \(1/a\) cancels the factor \(a\). Therefore the derivative of the proposed antiderivative is exactly the reciprocal integrand.
For a power term with \(r\ne-1\), differentiating \(x^{r+1}/(r+1)\) gives \(x^r\). Thus, after the algebraic identity has exposed powers and reciprocal terms, the familiar antiderivative formulas produce a function whose derivative equals the original expression.
Worked Example: Combining Several Basic Terms
Worked Example: Combining Several Basic Terms
Find an antiderivative of \(\displaystyle \frac{x^2+3x-2}{x^2}+\frac{4x+1}{2x-3}\), and check it.
The original expression is undefined at \(x=0\) and \(x=3/2\). First rewrite each fraction separately:
Adding these identities gives
Integrate the terms in order. The term \(-2x^{-2}\) has antiderivative \(2x^{-1}\), and the final term uses the linear reciprocal formula:
Differentiate and compare with the rewritten integrand:
This confirms the answer on each interval in the domain: \((-\infty,0)\), \((0,3/2)\), and \((3/2,\infty)\). The constant of integration can be chosen independently on those separate intervals.
Common Pitfall: Treating Every Negative Power Like a Reciprocal
A frequent error is to use a logarithm whenever a denominator contains \(x\). Only the power \(x^{-1}=1/x\) integrates to a logarithm. For a different negative power, use the power rule.
Correction: Rewrite \(5/x^2\) as \(5x^{-2}\), then integrate: $$ \int 5x^{-2}\,dx = 5\left(\frac{x^{-1}}{-1}\right)+C = -\frac{5}{x}+C. $$ Differentiation checks the correction: \(\displaystyle \frac{d}{dx}(-5/x)=5/x^2\).
Another useful check is to substitute the proposed decomposition back into the fraction, or to differentiate the final antiderivative. These checks catch missing factors, especially the factor \(1/a\) in an antiderivative involving \(ax+b\).
A Reliable Procedure
Find where the original denominator is zero. The antiderivative must be considered on intervals within the remaining domain.
Separate numerator terms over powers of \(x\), or divide a linear numerator by its linear denominator to obtain a constant and a reciprocal term.
Use the power rule unless the exponent is \(-1\); use a logarithm for reciprocal terms, including the required coefficient for a linear denominator.
Differentiate the proposed result and verify that it equals the original integrand wherever that integrand is defined.
This approach applies when simple algebra reveals powers and reciprocal terms. A more complicated rational expression may require additional methods; do not assume every rational function can be handled by these few rewrites alone.
Check Your Understanding
For each question, state any excluded values and show enough work to justify the result.
- Rewrite \(\displaystyle \frac{2x^2-5x+1}{x^2}\) as a sum of powers and reciprocal terms. Which term integrates to a logarithm?
- Find an antiderivative of \(\displaystyle \frac{6x+5}{3x+1}\). Check it by differentiation.
- Why does \(\displaystyle \int x^{-3}\,dx\) use the power rule rather than a logarithm?
- Find an antiderivative of \(\displaystyle \frac{4}{x^2}\), and identify the specific error in the claim that its antiderivative is \(4\ln|x|\).
- For \(\displaystyle \frac{3x+2}{x-4}\), rewrite the fraction as a constant plus a reciprocal term before integrating.
