Why Simplify Before Decomposing?
A rational integrand is a quotient of polynomials. Partial fractions can rewrite some such quotients as sums of simpler fractions, but setting up that decomposition takes work. If factoring, cancellation, or a substitution already makes the integral straightforward, decomposition adds steps without helping.
The useful habit is to inspect the whole integrand first. Look for common factors, compare the numerator with the derivative of the denominator, and check whether the denominator is a power of a simpler expression. These checks can reveal a direct route to an antiderivative.
Cancellation Is About Factors
A common factor is an expression that multiplies both the numerator and denominator. Where that factor is nonzero, it can be canceled. The restriction “where it is nonzero” matters: canceling a factor does not change the original expression’s domain.
For integration, this means the simplified expression can be integrated on intervals where the original integrand is defined. If cancellation removes a point from the simplified denominator, the original integrand may still be undefined at that point. Keep that restriction in mind when describing where an antiderivative applies.
Worked Example: Cancel a Common Factor First
Worked Example: Cancel a Common Factor First
Evaluate \(\displaystyle\int \frac{x^2-4}{x^2-x-2}\,dx\).
First factor the numerator and denominator:
The original integrand is undefined at \(x=2\) and \(x=-1\). Cancel the common factor \(x-2\), retaining those original restrictions:
Rewrite the numerator as \(x+2=(x+1)+1\), so the quotient becomes:
Integrate each term:
Check by differentiating. At every point in the original domain, \(x\ne 2,-1\):
The expression is an antiderivative on each interval in the original domain: \((-\infty,-1)\), \((-1,2)\), and \((2,\infty)\). The formula simplifies at \(x=2\), but the original integrand remains undefined there.
Why the Simplification Works
The key algebraic fact is that the canceled expression agrees with the original wherever the canceled factor is nonzero. In the example, multiplying out shows that both quotients have the same value at every allowed \(x\). Consequently, their derivatives and integrals can be compared on any interval that stays within the original domain.
More generally, suppose \(P(x)=A(x)R(x)\) and \(Q(x)=B(x)R(x)\). At every point where \(R(x)\ne0\) and \(B(x)\ne0\), division gives:
Integrating the simpler expression is therefore legitimate on those intervals. The caution about the domain follows from the same reasoning: this equality was established only where the canceled factor was nonzero.
Look for the Derivative of the Denominator
Another useful check is to differentiate the denominator. If the numerator is the same as that derivative, or differs only by a constant factor, substitution can turn the fraction into a logarithm. This is an application of the chain rule in reverse: differentiating \(\ln|g(x)|\) gives \(g'(x)/g(x)\) wherever \(g(x)\ne0\).
Worked Example: Match the Numerator to the Denominator's Derivative
Evaluate \(\displaystyle\int \frac{2x+3}{x^2+3x+5}\,dx\).
Let the denominator be \(g(x)=x^2+3x+5\). Its derivative is \(g'(x)=2x+3\), exactly the numerator. Set \(u=x^2+3x+5\). Then \(du=(2x+3)\,dx\), and the integral becomes:
To see that the logarithm's argument is positive, complete the square:
Thus the absolute value could be omitted, although retaining it in the antiderivative is also valid. Check by differentiation:
The derivative matches the integrand for every real \(x\), since the denominator is always positive.
Substitute When a Simpler Expression Is Raised to a Power
The numerator need not be the derivative of the denominator in exactly the same written form. What matters is whether a substitution absorbs the numerator and leaves a basic power to integrate. This is especially useful when a polynomial expression is raised to a power in the denominator.
Worked Example: Use Substitution Instead of Decomposition
Evaluate \(\displaystyle\int \frac{2x}{(x^2+1)^3}\,dx\).
Set \(u=x^2+1\), so \(du=2x\,dx\). The integral becomes a power of \(u\):
Check by differentiating the proposed antiderivative, using the chain rule:
The result checks exactly. Partial fractions would introduce extra algebra here, while substitution reduces the integral immediately to a single power.
When Is Partial Fractions Worthwhile?
Partial fractions are a way to rewrite a rational function as a sum of simpler rational functions, usually after factoring the denominator. They are useful when the resulting terms have standard antiderivatives that are easier to integrate than the original quotient. They are not a required first step for every rational integrand.
| What you notice | What to try first |
|---|---|
| Numerator and denominator share a polynomial factor | Factor both, then cancel the common factor while preserving the original restrictions. |
| The numerator is the derivative of the denominator, up to a constant | Substitute the denominator as the new variable. |
| A simple expression in \(x\) is raised to a power, with its derivative in the numerator | Substitute that expression and integrate the resulting power. |
| The fraction simplifies by a direct algebraic rewrite | Rewrite it before choosing a decomposition. |
| The denominator factors, and no cancellation or useful substitution simplifies the quotient | Partial fractions may be an efficient next method. |
This is a strategy guide, not a rule that one method always wins. A quick check of the factors and derivatives can save substantial work. If those checks do not simplify the integrand, partial fractions may be exactly the right tool.
Common Pitfall: Canceling Terms Instead of Factors
Cancellation is valid for common factors, not for individual terms in a sum. For example, starting with the simplified fraction from the first worked example, it would be incorrect to cancel the \(x\)'s in \((x+2)/(x+1)\) and claim that the result is \(2\). The \(x\)'s are terms in sums, not factors multiplying the entire numerator and denominator.
The correct calculation rewrites the numerator as \((x+1)+1\):
not \(2\). Integrating the incorrect result would give \(2x+C\), whose derivative is \(2\), not the original integrand. The corrected antiderivative is \(x+\ln|x+1|+C\), and its derivative is \(1+1/(x+1)=(x+2)/(x+1)\), as required.
Why This Strategy Matters
Integration methods can overlap: the same integral may be approachable in more than one way. Choosing a method that respects the structure already present in the integrand makes the calculation shorter and easier to check. For rational functions, the most useful first observations are often simple: a common factor disappears, a numerator matches a denominator's derivative, or a substitution turns a complicated-looking quotient into a basic power.
After simplifying, verify the result by differentiation. That check tests the antiderivative against the integrand that was actually given, including its algebra and signs. For a rational integrand, also remember where the original denominator was zero; cancellation does not restore those points to the original domain.
Check Your Understanding
Answer these questions, and check your reasoning by simplifying or differentiating where appropriate.
- What must be true of two expressions before a factor can be canceled from their quotient?
- If cancellation removes a point from the denominator, what should you remember about the original integrand's domain?
- What substitution is suggested when the numerator is the derivative of the denominator?
- For \(\displaystyle\int \frac{2x}{(x^2+1)^4}\,dx\), what substitution reduces the integrand to a power?
- Why is it invalid to cancel the \(x\)'s in \((x+2)/(x+1)\)?
- Name one situation in which partial fractions may still be useful.
