A rational function \dfrac{P(x)}{Q(x)} looks hopeless to integrate
as written. But if its denominator factors, the fraction can be split into a
sum of simpler fractions — each of which integrates to a logarithm by the
substitution
u = x - r. This decomposition is the method of
partial fractions.
\frac{1}{(x - r)(x - s)} = \frac{A}{x - r} + \frac{B}{x - s}.
Find the constants A, B, and the impossible integral becomes a sum
of \int \tfrac{dx}{x - r} = \ln|x - r| pieces.
Worked example: \displaystyle\int \frac{dx}{(x-1)(x+2)}
Step 1 — write the decomposition with unknowns. Two distinct linear factors,
so two simple fractions:
\frac{1}{(x-1)(x+2)} = \frac{A}{x-1} + \frac{B}{x+2}.
Step 2 — clear the denominators. Multiply both sides by
(x-1)(x+2):
1 = A(x+2) + B(x-1).
This must hold for every x, which lets us pick convenient
values.
Step 3 — cover-up at x = 1. Setting
x = 1 kills the B term
(x - 1 = 0):
1 = A(1 + 2) + B(0) = 3A \qquad\Longrightarrow\qquad A = \frac{1}{3}.
Step 4 — cover-up at x = -2. Setting
x = -2 kills the A term:
1 = A(0) + B(-2 - 1) = -3B \qquad\Longrightarrow\qquad B = -\frac{1}{3}.
Step 5 — write the decomposition.
\frac{1}{(x-1)(x+2)} = \frac{1/3}{x-1} - \frac{1/3}{x+2}.
Step 6 — integrate each piece. Each is a logarithm (via
u = x - 1 and u = x + 2):
\int \frac{dx}{(x-1)(x+2)} = \frac{1}{3}\ln|x-1| - \frac{1}{3}\ln|x+2| + C.
Equivalently, combining the logs,
\tfrac{1}{3}\ln\left|\dfrac{x-1}{x+2}\right| + C.
The shapes of decomposition
Distinct linear factors are the easy case; two more shapes round out the method. Each
repeated factor and each irreducible quadratic gets its own template of unknowns.
Write a proper rational function (degree of P below
degree of Q) as a sum of these building blocks, one per factor of
Q:
-
Distinct linear factor (x - r) contributes
\dfrac{A}{x - r}.
-
Repeated linear factor (x - r)^k contributes
one term per power,
\dfrac{A_1}{x - r} + \dfrac{A_2}{(x - r)^2} + \cdots + \dfrac{A_k}{(x - r)^k}.
-
Irreducible quadratic factor (x^2 + bx + c)
contributes \dfrac{Bx + C}{x^2 + bx + c} (a linear numerator).
Cover-up. For a distinct linear factor
(x - r), you can read its constant off instantly: cover that
factor in the denominator and evaluate what remains at x = r. For
our example, the constant over (x - 1) is
A = \left.\frac{1}{\;x + 2\;}\right|_{x = 1} = \frac{1}{3},
and the constant over (x + 2) is
\left.\tfrac{1}{x - 1}\right|_{x = -2} = -\tfrac{1}{3} — exactly
Steps 3–4, done in your head. (Cover-up shortcut applies only to simple linear factors; for
repeated or quadratic factors, equate coefficients instead.)
Make it proper first. Partial fractions require the numerator's degree to be
strictly less than the denominator's. If \deg P \ge \deg Q,
do polynomial long division
first to peel off a polynomial part, leaving a proper remainder fraction to decompose:
\frac{x^2}{(x-1)(x+2)} = 1 + \frac{-x + 2}{(x-1)(x+2)},
where the polynomial part 1 integrates trivially and the proper
remainder is decomposed as usual.