Partial Fractions and Integration Strategy

A rational function is a ratio $f(x)=P(x)/Q(x)$ of polynomials. Every such function has an elementary antiderivative, and there is a mechanical procedure to find it: rewrite $f$ as a sum of simpler fractions, each of which integrates to a logarithm or an arctangent. The rewriting is the method of partial fractions, running the addition of fractions in reverse.

The direction to reverse looks like this: combining two fractions over a common denominator gives

Read backward, the messy right side splits into two pieces that each integrate at sight:

Proper fractions and factoring

Partial fractions applies only to a proper rational function, one with $\mathrm{deg}P<\mathrm{deg}Q$. If $f$ is improper, divide first: polynomial long division writes

where the quotient $S$ integrates directly and the proper remainder $R(x)/Q(x)$ goes to partial fractions.

Once the fraction is proper, factor the denominator. Any polynomial with real coefficients factors completely into linear factors $ax+b$ and irreducible quadratic factors $a{x}^2+bx+c$ (those with $b^2-4ac<0$, which have no real root). The shape of that factorization determines the form of the decomposition, in four cases.

Case I: distinct linear factors

If $Q$ is a product of distinct linear factors, each contributes one term with an unknown constant on top. For $\int \frac{x^2+2x-1}{2x^3+3x^2-2x}\mathbf{d}x$, factor the denominator as $x(2x-1)(x+2)$ and write

Multiply through by $x(2x-1)(x+2)$:

The cover-up shortcut

This identity holds for every $x$, so choosing values at which factors vanish isolates one constant at a time. Setting $x=0$ leaves $-1=A(-1)(2)$, so $A=\frac{1}{2}$. Setting $x=\frac{1}{2}$ leaves only the $B$ term; setting $x=-2$ leaves only $C$:

Each root of a linear factor covers up that factor and reads its coefficient directly.

With the constants known,

Every distinct-linear case integrates to a sum of logarithms.

Case II: repeated linear factors

A factor $(ax+b)$ repeated $r$ times contributes one term for each power from $1$ up to $r$. For $\int \frac{x^4-2x^2+4x+1}{x^3-x^2-x+1}\mathbf{d}x$, the integrand is improper, so divide first:

The denominator factors as $(x-1{)}^2(x+1)$, so the repeated factor $(x-1)$ gets two terms:

Case III: irreducible quadratic factors

An irreducible quadratic $a{x}^2+bx+c$ contributes a term with a linear numerator, $(Ax+B)/(a{x}^2+bx+c)$. Such a term integrates to a logarithm plus an arctangent, using

For $\int \frac{2x^2-x+4}{x^3+4x}\mathbf{d}x$, the denominator $x(x^2+4)$ has one linear and one irreducible quadratic factor:

Matching coefficients gives $A=1$, $B=1$, $C=-1$. Splitting the quadratic term into a piece whose numerator is the derivative of $x^2+4$ and a constant piece,

When the quadratic does not factor and its numerator is not already a derivative, complete the square in the denominator first, then substitute to reach the arctangent form.

Case IV: repeated irreducible quadratic factors

A factor $(a{x}^2+bx+c{)}^r$ contributes one linear-over-quadratic term per power:

Each term integrates by substitution or by completing the square, though the higher powers grow tedious by hand and are usually left to a computer algebra system.

The four cases together are exhaustive: any proper rational function decomposes into terms of these forms, so any rational function has an elementary antiderivative.

Rationalizing substitutions

An integrand containing $\sqrt[n]{g(x)}$ can sometimes be made rational by substituting $u=\sqrt[n]{g(x)}$.

The overall strategy

Differentiation is mechanical: the rules say which formula applies. Integration is not. Facing an unfamiliar integral, work through four steps.¹

1. Simplify the integrand. Algebra or a trigonometric identity may make the method obvious, as when $\int (\sin x+\cos x{)}^2\mathbf{d}x$ expands to $\int (1+2\sin x\cos x)\mathbf{d}x$.
2. Look for an obvious substitution. Find a $u=g(x)$ whose differential $g^{\prime }(x)\mathbf{d}x$ is already present up to a constant.
3. Classify by the form of the integrand.
   • Trigonometric products of powers use the trigonometric-integral strategies.
   • Rational functions use partial fractions.
   • A power of $x$ times a transcendental function uses integration by parts.
   • A quadratic radical $\sqrt{\pm x^2\pm a^2}$ uses trigonometric substitution; $\sqrt[n]{ax+b}$ uses a rationalizing substitution.
4. Try again. Only two general methods exist, substitution and parts, so if the first three steps stall, attempt a less obvious substitution, apply parts to a single function, or manipulate the integrand toward a form already solved.

Nonelementary antiderivatives

The strategy does not always succeed. The elementary functions are those built from polynomials, roots, exponentials, logarithms, and trigonometric functions by finitely many arithmetic operations and compositions. Their derivatives are always elementary, but their integrals need not be. It has been proved that

have no elementary antiderivative. Each still defines a perfectly good function by the Fundamental Theorem, just not one expressible in closed form. Such integrals are handled by power series or by numerical approximation.

Tables and computer algebra

Extensive tables of integrals list hundreds of forms indexed by shape; matching a given integrand usually needs a preliminary substitution or algebraic step to fit one of the listed patterns. Computer algebra systems automate the same pattern-matching and handle the tedious cases (repeated quadratic factors, high-degree decompositions) instantly. Two cautions: a system returns a single antiderivative with no constant of integration, and it may drop absolute-value bars inside logarithms, valid only on part of the domain. A hand computation sometimes yields a cleaner form than the machine's.²