3.4 Partial Fractions

Big idea

You can integrate $\frac{1}{x-2}$ and you can integrate $\frac{1}{x+1}$, but handed $\frac{5x-4}{x^2-x-2}$ you have nothing to work with. Substitution fails because the numerator is not a multiple of the derivative of the denominator, and there is no product to split with parts. The obstruction is purely algebraic: the integrand has been added together, and addition is exactly what you need to undo.

Partial fraction decomposition reverses the step of putting fractions over a common denominator. Every rational function whose denominator factors over the reals can be written as a sum of simpler fractions whose denominators are those factors, and each simple fraction has an antiderivative you already know: a logarithm, a power, or an arctangent. So the calculus in this section is trivial once the algebra is done, and the section is really an algebra technique with an integral attached.

Two structural facts from algebra make the method work in every case. A polynomial with real coefficients factors completely into linear factors and irreducible quadratic factors, possibly repeated. And a rational function can be split into a polynomial plus a proper rational function, where proper means the numerator has strictly smaller degree than the denominator. Those two facts together say that the list of possible pieces is short and known in advance.

The order of operations matters. Check properness first, divide if you must, then factor the denominator, then write the template with unknown constants, then solve for them. Skipping the division step produces a decomposition that cannot exist, and you will discover this only after a page of algebra that refuses to close.

Decoder

A proper rational function is a sum of terms, one for each power of each factor of its denominator: a constant over each power of a linear factor, and a linear expression over each power of an irreducible quadratic factor.

The word doing the work is proper. If the degree of the numerator is greater than or equal to the degree of the denominator, no such sum exists, because every term on the right is proper and a sum of proper fractions is proper. Polynomial long division fixes this by peeling off a polynomial quotient, and the remainder over the divisor is proper by construction.

The second phrase to read carefully is each power of each factor. A factor $(x+2)^2$ does not contribute one term; it contributes two, one with $(x+2)$ in the denominator and one with $(x+2)^2$. The reason is a counting argument: the number of unknown constants must equal the degree of the denominator, or the system that determines them is not solvable for every numerator.

The classic mistake is putting a constant over an irreducible quadratic. A term like $\frac{A}{x^2+4}$ cannot represent a general numerator; you need $\frac{Ax+B}{x^2+4}$, which contributes the two constants that the quadratic factor’s degree demands.

Definitions and results

Proper and improper. A rational function $\frac{P(x)}{Q(x)}$ is proper when $\deg P < \deg Q$. If it is not, divide: $\frac{P}{Q} = S + \frac{R}{Q}$ with $\deg R < \deg Q$, integrate the polynomial $S$ directly, and decompose $\frac{R}{Q}$.

Distinct linear factors. If $Q$ factors as a product of distinct linear factors $(x-r_1)\cdots(x-r_n)$, the decomposition is

$$ \frac{P(x)}{Q(x)} = \frac{A_1}{x-r_1} + \cdots + \frac{A_n}{x-r_n} $$

and each term integrates to $A_i\ln|x-r_i|$.

Repeated linear factors. A factor $(x-r)^k$ contributes $k$ terms:

$$ \frac{A_1}{x-r} + \frac{A_2}{(x-r)^2} + \cdots + \frac{A_k}{(x-r)^k} $$

Only the first term produces a logarithm; the others integrate to negative powers.

Irreducible quadratic factors. A quadratic $x^2+bx+c$ with $b^2 - 4c < 0$ cannot be factored over the reals. Each such factor contributes a term $\frac{Ax+B}{x^2+bx+c}$, and a repeated one $(x^2+bx+c)^k$ contributes $k$ such terms with increasing powers in the denominators.

Solving for the constants. Multiply both sides by $Q(x)$ to clear denominators, producing an identity between polynomials. Either substitute the roots of the linear factors to isolate constants one at a time, or expand and match coefficients of like powers. The two methods agree; the first is faster when the roots are real and simple.

The three antiderivatives you land on. Every piece reduces to one of

$$ \int\frac{dx}{x-r} = \ln|x-r| + C, \quad \int\frac{dx}{(x-r)^k} = \frac{(x-r)^{1-k}}{1-k} + C, \quad \int\frac{dx}{x^2+a^2} = \frac{1}{a}\arctan\frac{x}{a} + C $$

Handling the quadratic term. Split $\frac{Ax+B}{x^2+a^2}$ into a piece whose numerator is a multiple of the derivative of the denominator, which gives a logarithm by substitution, plus a constant over the quadratic, which gives an arctangent. If the quadratic has a linear term, complete the square first.

Worked examples

Two distinct linear factors

Evaluate $\int \frac{5x-4}{x^2-x-2}\,dx$.

The integrand is proper and the denominator factors as $(x-2)(x+1)$, so write

$$ \frac{5x-4}{(x-2)(x+1)} = \frac{A}{x-2} + \frac{B}{x+1} $$

Clearing denominators gives $5x - 4 = A(x+1) + B(x-2)$. Substituting $x = 2$ gives $6 = 3A$, so $A = 2$. Substituting $x = -1$ gives $-9 = -3B$, so $B = 3$. Therefore

$$ \int \frac{5x-4}{x^2-x-2}\,dx = 2\ln|x-2| + 3\ln|x+1| + C $$

Check the algebra by recombining: $\frac{2}{x-2} + \frac{3}{x+1} = \frac{2(x+1) + 3(x-2)}{(x-2)(x+1)} = \frac{5x-4}{x^2-x-2}$.

Dividing before decomposing

Evaluate $\int \frac{x^3}{x^2-1}\,dx$.

The numerator has degree $3$ and the denominator degree $2$, so the fraction is improper and no decomposition exists yet. Long division gives $x^3 = x(x^2-1) + x$, so

$$ \frac{x^3}{x^2-1} = x + \frac{x}{x^2-1} $$

The leftover is proper. Here the numerator is half the derivative of the denominator, so a substitution $u = x^2-1$ finishes it without any decomposition at all:

$$ \int \frac{x^3}{x^2-1}\,dx = \frac{x^2}{2} + \frac{1}{2}\ln\big|x^2-1\big| + C $$

This is worth noticing: after division, always look for a substitution before setting up unknowns.

A repeated linear factor

Evaluate $\int \frac{x+4}{x(x+2)^2}\,dx$.

The denominator has a simple factor $x$ and a repeated factor $(x+2)^2$, so the template has three terms:

$$ \frac{x+4}{x(x+2)^2} = \frac{A}{x} + \frac{B}{x+2} + \frac{C}{(x+2)^2} $$

Clearing denominators gives $x + 4 = A(x+2)^2 + Bx(x+2) + Cx$. Setting $x = 0$ gives $4 = 4A$, so $A = 1$. Setting $x = -2$ gives $2 = -2C$, so $C = -1$. For $B$, compare the coefficients of $x^2$: the left side has none, and the right side has $A + B$, so $B = -1$. Integrating term by term,

$$ \int \frac{x+4}{x(x+2)^2}\,dx = \ln|x| - \ln|x+2| + \frac{1}{x+2} + C $$

The middle two terms combine into $\ln\left|\frac{x}{x+2}\right|$ if you prefer a single logarithm.

An irreducible quadratic factor

Evaluate $\int \frac{2x^2-x+4}{x(x^2+4)}\,dx$.

The quadratic $x^2+4$ has no real roots, so it stays intact and carries a linear numerator:

$$ \frac{2x^2-x+4}{x(x^2+4)} = \frac{A}{x} + \frac{Bx + C}{x^2+4} $$

Clearing gives $2x^2 - x + 4 = A(x^2+4) + (Bx+C)x$. Setting $x = 0$ gives $4 = 4A$, so $A = 1$. Matching $x^2$ coefficients gives $2 = A + B$, so $B = 1$. Matching $x$ coefficients gives $-1 = C$. The integral splits into three standard pieces:

$$ \int\left(\frac{1}{x} + \frac{x}{x^2+4} - \frac{1}{x^2+4}\right)dx = \ln|x| + \frac{1}{2}\ln\big(x^2+4\big) - \frac{1}{2}\arctan\frac{x}{2} + C $$

Note how the linear numerator produced exactly the two kinds of term promised: the $x$ part became a logarithm by substitution, and the constant part became an arctangent.

Practice

These problems run the full pipeline: check whether division is needed, factor the denominator, write the correct template, solve for the constants, and integrate. Watch the factor types, since the template is determined entirely by them.

Quiz

Five items on properness and long division, decomposition templates for repeated and irreducible factors, and the resulting antiderivatives.