Partial Fractions and Algebraic Decomposition

Overview

Central question: Why split a rational expression before integrating? You will turn one complicated rational function into a sum of simple fractions, solve for the unknown coefficients by an identity, and verify the decomposition before taking any antiderivatives. The method is algebra first, calculus second.

Consider

$$\int\frac{3x+5}{(x-1)(x+2)}\,dx.$$

The denominator factors into two distinct linear factors, while the numerator has lower degree than the denominator. That makes the fraction proper and identifies the decomposition template. Before computing, predict the two simple denominators that must appear in the final sum.

Build the template from the factors

For distinct linear factors $(x-1)(x+2)$, write

$$\frac{3x+5}{(x-1)(x+2)} =\frac{A}{x-1}+\frac{B}{x+2}.$$

The symbols $A$ and $B$ are constants, not functions of $x$. Multiplying every term by the common denominator produces an identity:

$$3x+5=A(x+2)+B(x-1).$$

This equation must be true for every $x$ except the original excluded values. Convenient choices make one term vanish. At $x=1$, $8=3A$, so $A=\frac83$. At $x=-2$, $-1=-3B$, so $B=\frac13$.

The integral is now a sum of logarithmic forms:

$$\int\frac{3x+5}{(x-1)(x+2)}\,dx =\frac83\ln|x-1|+\frac13\ln|x+2|+C.$$

Absolute values matter because an antiderivative is valid on any interval that avoids the excluded values $x=1$ and $x=-2$. The original rational function is not defined at those points, so one expression cannot describe a differentiable antiderivative across them.

Verify the algebra before the calculus

Substitute the solved coefficients back into the right side:

$$\frac{\frac83}{x-1}+\frac{\frac13}{x+2} =\frac{8(x+2)+(x-1)}{3(x-1)(x+2)} =\frac{3x+5}{(x-1)(x+2)}.$$

This check is short and catches the most common error: solving the coefficient equations incorrectly and then integrating a wrong decomposition flawlessly. Differentiating the final logarithmic answer supplies a second, independent check.

Conditions before decomposition

If the numerator degree is at least the denominator degree, perform polynomial division first. For example, $\frac{x^2+1}{x-1}$ is not a proper rational function, so the template above is not the first move. Division gives $x+1+\frac2{x-1}$, which is immediately integrable.

If a denominator does not factor over the reals, its numerator in the decomposition must be linear rather than constant. Repeated factors require a term for every power. Those cases are not exceptions to memorize; they are consequences of giving the decomposition enough adjustable coefficients to reproduce every possible numerator.

Solve coefficients by comparing structure

Plugging in roots of denominator factors is efficient, but it is not the only legitimate coefficient method. Expand the identity

$$3x+5=A(x+2)+B(x-1)$$

to obtain

$$3x+5=(A+B)x+(2A-B).$$

Matching coefficients of equal powers of $x$ gives the linear system $A+B=3$ and $2A-B=5$. Solving produces $A=\frac83$ and $B=\frac13$, the same values found by input substitution. This method remains available when no convenient input value isolates a coefficient, such as a decomposition involving an irreducible quadratic factor.

Coefficient comparison has a conceptual advantage: it explains why partial fractions are an identity problem. The left and right sides are polynomials that agree for every allowed input. Corresponding coefficients must agree. Test values are a shortcut for solving that identity, not a separate rule that works by magic.

Domain and logarithms

The factors $(x-1)$ and $(x+2)$ divide the real line into three intervals. On each interval, the original rational function is continuous and has an antiderivative. The absolute values in $\ln|x-1|$ and $\ln|x+2|$ allow one formula to differentiate correctly on each such interval, but they do not make the original function defined at $x=1$ or $x=-2$.

When reporting a definite integral of a rational function, first check that the interval does not cross a denominator zero. If it does, the integral is improper and must be defined with limits; the logarithmic antiderivative alone does not authorize direct endpoint substitution across a vertical asymptote. This condition check connects algebraic decomposition to the limit reasoning developed later in the course.

A transfer setup

Set up, but do not solve, the decomposition of

$$\frac{x+4}{(x-2)(x^2+1)}.$$

The correct template is

$$\frac{A}{x-2}+\frac{Bx+C}{x^2+1}.$$

The numerator $Bx+C$ is linear because $x^2+1$ is an irreducible quadratic over the reals. A constant numerator would not give enough freedom to reproduce a general linear numerator after denominators are cleared. The next lesson develops this template logic for repeated factors and irreducible quadratics in full.

Stop and check

An incorrect decomposition is $\frac{A}{(x-1)(x+2)}+\frac{B}{(x-1)(x+2)}$. It leaves the same difficult denominator in both terms and cannot represent two independently adjustable simple fractions. The misconception is seeing partial fractions as adding labels rather than splitting factor structure. The correction is to give each denominator factor its own fraction.

Retrieval check

Set up the decomposition for $\frac{5x-1}{(x+3)(x-4)}$. Write the two unknown fractions, clear denominators, and select two input values that isolate the coefficients. Explain why $x=-3$ and $x=4$ are useful choices.

The next lesson expands the template logic to repeated and irreducible factors, where omitting a necessary term makes the coefficient identity impossible to satisfy.