Using Linear Partial Fractions

3. Visual Connection

5. Concept Development

Linear partial fractions is an AP Calculus BC technique

The method rewrites certain rational functions as sums of simpler ratios whose antiderivatives are logarithms. Topic 6.12 focuses on distinct, nonrepeating linear factors.

The rational function must be proper before decomposition

Check that \(\deg P<\deg Q\) in \(P(x)/Q(x)\). If the numerator degree is at least the denominator degree, perform polynomial long division first.

Factor the denominator completely over the reals

Identify every distinct linear factor before writing the decomposition. The factorization determines both the excluded domain values and the required fraction templates.

Use one constant numerator for each distinct linear factor

With three distinct factors, include three fractions; do not omit a factor or use a variable numerator over a linear denominator.

Clear denominators to create a polynomial identity

This identity must hold for every permissible \(x\), which allows the constants to be determined efficiently.

Strategic substitution isolates coefficients

Substitute a root of one denominator factor into the cleared identity. All terms containing that factor vanish, often leaving a one-step equation for one coefficient.

Equating coefficients is an alternative method

Expand the cleared identity and match coefficients of equal powers of \(x\). This is useful as a check or when strategic values do not determine every constant immediately.

The number of constants must match the number of factors

For \(n\) distinct linear factors, write \(n\) partial fractions and solve for \(n\) constants. A missing term makes recombination impossible in general.

Recombine before integrating

Place the proposed fractions over their common denominator. The resulting numerator must exactly equal \(P(x)\); this catches most sign and arithmetic errors.

Each linear fraction integrates to a logarithm

The factor \(1/m\) is required when the linear denominator is not monic.

Use one overall constant of integration

Integrate every decomposed term and add a single \(+C\) to the final sum. Separate constants from individual logarithms combine into that one arbitrary constant.

Logarithmic absolute values preserve interval validity

Each \(\ln|mx+b|\) is valid on intervals that do not cross its zero. The original rational function and its antiderivative family share those excluded inputs.

Logarithms may be combined after integration

Expressions such as \(A\ln|x-a|-A\ln|x-b|\) may be written as \(A\ln|(x-a)/(x-b)|\). Keeping them separate often makes verification easier.

Improper rational functions require division first

After long division, decompose only the proper remainder. The final antiderivative may contain a polynomial term together with logarithms.

Definite integrals require a domain check

Before endpoint evaluation, locate all denominator zeros. If one lies inside the interval, the integral is improper and must be split into one-sided limits; ordinary cancellation of divergent logarithms is invalid.

Know the scope boundary of this AP topic

Repeated linear factors and irreducible quadratic factors require different decomposition templates. They are useful broader-calculus ideas, but the required Topic 6.12 form uses ratios of distinct linear, nonrepeating factors.

A reliable decomposition workflow

Compare degrees, divide if necessary, factor the denominator, write one fraction per distinct factor, clear denominators, solve constants, recombine, integrate, and check the domain.

Common errors

Frequent errors include decomposing before division, incomplete factoring, omitting a fraction, losing a sign during strategic substitution, forgetting the linear coefficient in a logarithm, dropping absolute values, and evaluating across a pole.

6. Detailed Worked Example and Error Check

Example 1: Decompose two distinct linear factors.

Recombination gives \(2(x+2)+3(x-1)=5x+1\), so

Example 2: Use strategic substitution with a constant numerator.

Thus the integral is \(\frac15\ln|x-2|-\frac15\ln|x+3|+C\).

Example 3: Decompose three linear factors.

Therefore the antiderivative is \(\ln|x|-2\ln|x-1|+3\ln|x+2|+C\).

Example 4: Account for nonmonic linear factors.

Integration gives

Example 5: Divide before decomposing.

The antiderivative is \(x+2\ln|x-2|-\ln|x+1|+C\).

Example 6: Evaluate a proper definite integral.

Example 7: Detect an interior pole. Although \(1/[(x-1)(x+2)]=\frac13/(x-1)-\frac13/(x+2)\), the integral from 0 to 3 has a pole at \(x=1\). It must be split into one-sided improper integrals, and those limits do not both converge.