Integration using partial fractions: VCE Specialist Mathematics Units 3 & 4

1. Decompose the algebra before integrating

A rational function is a quotient of polynomials. Its numerator and denominator may conceal a sum of simpler fractions whose antiderivatives are already known. Partial fractions reverse the familiar process of combining fractions over a common denominator. The aim is not a new integration formula for every quotient: it is an algebraic identity that turns one difficult-looking integral into several standard ones.

Start by recording the original domain and factorising the denominator over the real numbers as far as useful. A factor that cancels still leaves an excluded value in the original function. If the numerator’s degree is at least the denominator’s degree, divide first. Partial fractions are then applied to the proper remainder, whose numerator has smaller degree than its denominator.

For example, x²/(x² + 1) = 1 − 1/(x² + 1). The integral immediately becomes x − arctan x + C. Attempting only constant numerators over the original denominator would miss the polynomial part. Polynomial division is not optional algebraic decoration; it exposes the portion of the function that does not decay like a proper fraction.

2. The factor structure chooses the template

A distinct linear factor x − a needs a constant numerator A. A repeated factor (x − a)² needs both A/(x − a) and B/(x − a)². An irreducible real quadratic such as x² + 4 needs a linear numerator Bx + C. These templates give enough flexibility to reproduce the original numerator after multiplication by the denominator.

For a repeated factor of order three, include all three powers of that factor. Leaving out an intermediate power can make the coefficient equations inconsistent even though a valid decomposition exists. For an irreducible quadratic, a constant numerator alone is too restrictive unless the solved coefficient of x happens to be zero. Choose the complete template before solving.

Multiply the proposed identity by the common denominator, then compare coefficients of equal powers of x. Convenient substitutions at roots of linear factors often find some coefficients quickly. These substitutions are made in the polynomial identity after denominators have been cleared; they do not claim the original rational function is defined at a pole. Use coefficient comparison or additional safe x-values for the remaining unknowns.

Finally recombine the fractions. This check is separate from integrating them. A correct antiderivative of an incorrect decomposition is still wrong. Checking the algebra first also makes it easier to locate errors: a coefficient problem belongs to the decomposition, while a missing logarithmic factor belongs to the integration.

3. Integrate each fraction according to its power

Only a first power of a linear factor produces a logarithm. A term A/(x − a)² is A(x − a)⁻² and integrates to −A/(x − a). Treating every partial fraction as logarithmic is a common error. If the factor is ax + b, include the reverse chain-rule factor 1/a in either the logarithmic or power result.

A quadratic numerator can be split into a multiple of the denominator’s derivative and a constant remainder. The derivative part gives a logarithm. Complete the square in the remaining denominator to expose an arctangent form. For instance, 4x + 7 = 2(2x + 2) + 3 matches the derivative of x² + 2x + 5, while that denominator equals (x + 1)² + 4.

The quadratic is positive everywhere in that example, so its logarithm needs no absolute-value bars. By contrast, linear factors change sign across their roots, and their logarithms need absolute values unless the interval already establishes a positive argument. The final answer inherits every original restriction, including any removable holes retained from earlier cancellation.

4. Definite integrals need a domain check first

Once an antiderivative has been found, definite evaluation is usually straightforward, provided the original integrand is continuous throughout the closed interval. Check every denominator zero against the whole interval, not just the two endpoints. A graph can help locate suspicious behaviour, but exact factorisation is better evidence when available.

For example, substituting −1 and 1 into ln|x| would give zero, yet that does not evaluate an ordinary integral of 1/x across zero. The integrand is undefined inside the interval, and a single application of the fundamental theorem is invalid. Recognising this obstruction is sufficient here; do not invent a cancellation rule across a pole.

When both bounds lie safely on one branch, logarithms can often be combined into exact ratios. Keep them separate until their signs and arguments are clear. Numerical approximations should be made only at the end. A CAS may display a different logarithmic combination; differentiation and a domain-aware simplification are the reliable equivalence tests.

In a rate model, poles also warn about the range of the model. A denominator becoming zero at a future time usually means the formula cannot simply be continued through that time. Integrating an algebraic expression is not permission to extend its physical meaning. State the time window used and interpret the accumulated quantity in the rate’s units multiplied by time.

5. An efficient examination and technology workflow

Use a predictable sequence: exclusions, division if needed, denominator factorisation, complete template, coefficient solution, recombination, integration, then any bounds or initial condition. This produces a readable argument and creates checkpoints before mistakes spread. A question that says “show that” needs the polynomial identity or equivalent reasoning, not just a calculator’s final decomposition.

When technology is permitted, use its partial-fraction command to check coefficients, then differentiate your own antiderivative. If the derivative simplifies to zero after subtracting the original integrand, still retain the original domain. Algebra software may cancel factors and hide holes. Plotting on either side of each excluded value can reveal their behaviour, but a finite plot cannot prove continuity.

To think beyond the routine, ask whether a parameter creates a repeated factor or cancels a pole. Such a change can alter both the correct decomposition template and the allowed intervals. Analyse the special parameter value separately rather than substituting into a formula derived by division by that parameter. This is a useful small investigation in algebra and calculus, without needing any advanced convergence theory.

6. Notice when a parameter changes the template

Consider the family 1/[x(x − a)]. If a ≠ 0, the factors are distinct and the identity is −1/(ax) + 1/[a(x − a)]. Clearing denominators gives −(x − a)/a + x/a = 1, so the coefficients are justified. Integration then gives (1/a)[ln|x − a| − ln|x|] + C on any interval avoiding 0 and a.

If a = 0, that formula cannot be used because its coefficients divide by a. The original integrand instead becomes 1/x², whose antiderivative is −1/x + C. The two linear factors have merged into a repeated factor, changing the appropriate template. This is a simple example of why special parameter values need separate reasoning.

A graph supplies a complementary interpretation: for nonzero a there are two potential pole locations, whereas at a = 0 they coincide. The correct algebra must reflect this changed geometry. In a linked question, establish the parameter condition before presenting a general expression, and retain the original variable exclusions in every case.

When checking a computer-generated result, inspect any denominators involving parameters as carefully as denominators involving x. A symbolic output can be valid only under a hidden nonzero assumption. Recombining the fractions with the parameter left symbolic tests the general identity, while substituting the exceptional value into the original expression tells you how to handle the separate case.