Rational Integration and Partial Fractions

Prepare the algebra before integrating

A rational integrand is a ratio of polynomials. If the numerator degree is at least the denominator degree, first perform polynomial division. Factor the remaining denominator over the reals. The form of the partial-fraction decomposition follows those factors, not a guessed list of simple fractions.

A repeated linear factor $(x-a)^m$ needs terms for every power from $1$ through $m$ . An irreducible quadratic needs a linear numerator, and repeated quadratics need one such numerator over every power. Solve for the coefficients by multiplying through by the common denominator and matching polynomial identities.

Visual guide

VISUAL GUIDEA rational function as two simpler pieces

1/x
1/(x + 1)
Difference

For x > 0, 1/[x(x + 1)] = 1/x − 1/(x + 1). The vertical gap between the two reciprocal curves equals the rational integrand. Decomposition turns its antiderivative into a difference of logarithms.

Worked example: two distinct factors

For $(3x+5)/[(x+1)(x+2)]$ , write

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

Multiplying through gives $3x+5=A(x+2)+B(x+1)$ . At $x=-1$ , $A=2$ ; at $x=-2$ , $B=1$ . Therefore the integral is $2\ln|x+1|+\ln|x+2|+C$ on each interval excluding the poles.

Using a pole value while solving the polynomial identity is valid: after clearing denominators, the identity is a polynomial statement. It does not make the original rational function defined at that pole.

Worked example: a repeated factor

Decompose $1/[x(x+1)^2]$ as $A/x+B/(x+1)+C/(x+1)^2$ . Matching coefficients gives $A=1$ , $B=-1$ , $C=-1$ . Integrating yields

\ln|x|-\ln|x+1|+\frac1{x+1}+C_0.

The squared denominator integrates by a power rule, not a logarithm. Differentiating this expression verifies the repeated-factor term's sign.

Quadratic denominators and method selection

For $(2x+3)/(x^2+1)$ , split the numerator into the derivative of the denominator and a remainder. The integral is $\ln(x^2+1)+3\arctan x+C$ . More generally, complete the square to handle an irreducible quadratic after separating any derivative-of-denominator piece.

A rational function may be integrated more quickly by a direct substitution or polynomial division. Check for those before setting up a large coefficient system. Regardless of the method, preserve excluded inputs and separate improper integrals at any interior pole.

Practice

Integrate $1/[x(x+1)]$ .
Integrate $(x^2+1)/(x-1)$ .
What numerator is required over an irreducible quadratic factor $x^2+x+1$ ?

Show worked solutions

The decomposition is $1/x-1/(x+1)$ , giving $\ln|x|-\ln|x+1|+C$ .
Divide first: $x+1+2/(x-1)$ . Integrate to $x^2/2+x+2\ln|x-1|+C$ .
Use $Ax+B$ . A constant alone may not be able to represent the original rational expression.