Partial fraction decomposition splits a complicated rational function into a sum of simple fractions that you already know how to integrate, mostly into logs and arctangents.
$$\frac{3x + 5}{(x+1)(x+3)} = \frac{1}{x+1} + \frac{2}{x+3}$$
It is the standard method for integrating any rational function, that is, any polynomial divided by another polynomial. Once you know it, every such integral becomes routine.
Why it works
Partial fractions is ordinary fraction addition run in reverse. If you add \( \frac{1}{x+1} + \frac{2}{x+3} \) over the common denominator \( (x+1)(x+3) \), the numerator is \( (x+3) + 2(x+1) = 3x + 5 \). Decomposition asks the opposite question: given the combined fraction, which simple pieces produced it?
A theorem from algebra guarantees the answer always exists and is unique. Every real polynomial factors into linear factors and irreducible quadratics, and every proper rational function can be written as a sum of simple fractions over those factors. The number of unknown constants always equals the degree of the denominator, so you get exactly as many equations as unknowns.
The payoff for integration is that each simple piece has a known antiderivative: \( \frac{A}{x-a} \) integrates to a logarithm (see integral of 1/x), \( \frac{A}{(x-a)^k} \) with \( k \ge 2 \) integrates to a power, and quadratic pieces give a log plus an arctangent.
Before you start: two checks
- Is the fraction proper? The degree of the top must be less than the degree of the bottom. If not, do polynomial long division first.
- Factor the denominator completely into linear factors and irreducible quadratics.
The method in five steps
- Divide if the fraction is improper, and set the polynomial part aside.
- Factor the denominator completely.
- Write the form: one term per linear factor, one term per power of a repeated factor, and a linear numerator over each irreducible quadratic.
- Solve for the constants by multiplying through by the denominator, then plugging in roots and comparing coefficients.
- Integrate each piece and add the results.
Before integrating, it pays to check the constants. Pick any convenient \( x \) that is not a root, such as \( x = 1 \) or \( x = 2 \), and confirm that the original fraction and your sum of pieces give the same number. It takes ten seconds and catches almost every arithmetic slip.
Case 1: distinct linear factors
Each factor \( (x - a) \) gets a term \( \frac{A}{x - a} \).
Example: \( \int\frac{3x+5}{(x+1)(x+3)}\,dx \).
$$\frac{3x+5}{(x+1)(x+3)} = \frac{A}{x+1} + \frac{B}{x+3} \quad\Longrightarrow\quad 3x + 5 = A(x+3) + B(x+1)$$
Cover-up trick: set \( x = -1 \): \( 2 = 2A \), so \( A = 1 \). Set \( x = -3 \): \( -4 = -2B \), so \( B = 2 \). Then
$$\int\left(\frac{1}{x+1} + \frac{2}{x+3}\right)dx = \ln|x+1| + 2\ln|x+3| + C$$
Quick check at \( x = 1 \): the original fraction is \( \frac{8}{2\cdot4} = 1 \), and the pieces give \( \frac12 + \frac24 = 1 \). The constants are right.
Why is it legal to plug in \( x = -1 \) when the original fraction is undefined there? Because after clearing denominators, \( 3x + 5 = A(x+3) + B(x+1) \) is an identity between two polynomials. Two polynomials that agree for all \( x \) other than \( -1 \) and \( -3 \) must agree everywhere, including at those points.
Classic: \( \frac{1}{x^2 - 1} = \frac{1/2}{x-1} - \frac{1/2}{x+1} \), so
$$\int\frac{dx}{x^2-1} = \frac12\ln\left|\frac{x-1}{x+1}\right| + C, \qquad \int_2^3\frac{dx}{x^2-1} = \frac12\ln\frac32$$
Example: factor first. For \( \int\frac{3\,dx}{x^2 + x - 2} \), first factor the denominator as \( (x-1)(x+2) \). Cover-up at \( x = 1 \) gives \( A = \frac{3}{3} = 1 \), and at \( x = -2 \) gives \( B = \frac{3}{-3} = -1 \). So the integral is \( \ln|x-1| - \ln|x+2| + C \).
Case 2: repeated linear factors
A factor \( (x - a)^k \) needs \( k \) terms: \( \frac{A_1}{x-a} + \frac{A_2}{(x-a)^2} + \dots + \frac{A_k}{(x-a)^k} \).
Example: \( \frac{1}{x(x+1)^2} = \frac{A}{x} + \frac{B}{x+1} + \frac{C}{(x+1)^2} \).
Clearing denominators: \( 1 = A(x+1)^2 + Bx(x+1) + Cx \). With \( x = 0 \): \( A = 1 \). With \( x = -1 \): \( C = -1 \). Comparing \( x^2 \) coefficients: \( A + B = 0 \), so \( B = -1 \).
$$\int\frac{dx}{x(x+1)^2} = \ln|x| - \ln|x+1| + \frac{1}{x+1} + C$$
Notice that the plug-in trick found two of the three constants, and comparing one coefficient found the last. Mixing the two methods is usually the fastest route.
Case 3: irreducible quadratic factors
A factor like \( x^2 + 1 \) gets a linear numerator: \( \frac{Bx + C}{x^2+1} \).
Example: \( \frac{1}{x(x^2+1)} = \frac{A}{x} + \frac{Bx + C}{x^2+1} \). Clearing denominators gives \( 1 = A(x^2+1) + (Bx + C)x \). Setting \( x = 0 \) kills the second term and leaves \( A = 1 \). There is no other real root to plug in, so compare coefficients instead: the \( x^2 \) terms give \( 0 = A + B \), so \( B = -1 \), and the \( x \) terms give \( 0 = C \). With \( A = 1,\ B = -1,\ C = 0 \):
$$\int\frac{dx}{x(x^2+1)} = \ln|x| - \frac12\ln(x^2+1) + C$$
Quadratic pieces integrate to a log (via u-substitution) plus an arctangent (see derivative of arctan x).
Example with an arctangent. Decompose \( \frac{2x^2 + 2x + 2}{(x+1)(x^2+1)} = \frac{A}{x+1} + \frac{Bx + C}{x^2+1} \). Clearing denominators,
$$2x^2 + 2x + 2 = A(x^2+1) + (Bx + C)(x+1)$$
At \( x = -1 \): \( 2 = 2A \), so \( A = 1 \). Comparing \( x^2 \) coefficients, \( 2 = A + B \), so \( B = 1 \). Comparing constants, \( 2 = A + C \), so \( C = 1 \). The quadratic piece \( \frac{x + 1}{x^2+1} \) splits into \( \frac{x}{x^2+1} \), which gives a log, and \( \frac{1}{x^2+1} \), which gives an arctangent:
$$\int\frac{2x^2+2x+2}{(x+1)(x^2+1)}\,dx = \ln|x+1| + \frac12\ln(x^2+1) + \arctan x + C$$
Completing the square. If the irreducible quadratic is not of the form \( x^2 + a^2 \), complete the square first. For example \( x^2 + 2x + 5 = (x+1)^2 + 4 \), so
$$\int\frac{dx}{x^2+2x+5} = \frac12\arctan\frac{x+1}{2} + C$$
When the fraction is improper: divide first
For \( \int\frac{x^2}{x^2-1}\,dx \), the degrees are equal, so divide: \( \frac{x^2}{x^2-1} = 1 + \frac{1}{x^2-1} \). The polynomial part integrates directly and the remainder is the classic example above:
$$\int\frac{x^2}{x^2-1}\,dx = x + \frac12\ln\left|\frac{x-1}{x+1}\right| + C$$
Sometimes the remainder is simpler than you expect. Dividing \( x^3 + x \) by \( x^2 - 1 \) gives \( x + \frac{2x}{x^2-1} \), and the remainder’s numerator is exactly the derivative of its denominator, so \( \int\frac{x^3+x}{x^2-1}dx = \frac{x^2}{2} + \ln|x^2-1| + C \) with no decomposition needed at all.
Summary table
| Denominator factor | Partial fraction terms |
|---|---|
| \( (x - a) \) | \( \frac{A}{x-a} \) |
| \( (x - a)^2 \) | \( \frac{A}{x-a} + \frac{B}{(x-a)^2} \) |
| \( x^2 + bx + c \) (irreducible) | \( \frac{Bx + C}{x^2 + bx + c} \) |
Common mistakes
- Skipping the division step. Decomposing \( \frac{x^2}{x^2-1} \) as \( \frac{A}{x-1} + \frac{B}{x+1} \) cannot work, because the right side is always proper. Divide first.
- Using one term for a repeated factor. \( (x+1)^2 \) needs both \( \frac{B}{x+1} \) and \( \frac{C}{(x+1)^2} \), not just one of them.
- A constant numerator over a quadratic. Over \( x^2 + 1 \) you need \( Bx + C \), not just \( C \). Using too few unknowns makes the system unsolvable.
- Treating a reducible quadratic as irreducible. \( x^2 - 4 = (x-2)(x+2) \) factors, so it gets two linear terms. Check the discriminant: \( x^2 + bx + c \) is irreducible only when \( b^2 - 4c < 0 \).
- Losing absolute values. \( \int\frac{dx}{x-a} = \ln|x - a| + C \), with the bars.
Where it’s used
Partial fractions is the reason the integral of sec x can be derived from scratch, since one approach rewrites it as \( \int\frac{\cos x}{1 - \sin^2 x}dx \). It appears in improper integrals of rational functions, in solving the logistic differential equation, and in telescoping series: because \( \frac{1}{n(n+1)} = \frac1n - \frac{1}{n+1} \), the sum \( \frac{1}{1\cdot2} + \frac{1}{2\cdot3} + \frac{1}{3\cdot4} + \frac{1}{4\cdot5} \) collapses to \( 1 - \frac15 \). Engineers use the same decomposition to invert Laplace transforms.
Check a definite integral
The integral calculator shows the full decomposition for indefinite integrals, and the definite integral calculator evaluates them between limits.
Practice problems
Try these before checking the answers.
- \( \int\frac{dx}{x(x - 2)} \)
- \( \int\frac{5x - 1}{(x-1)(x+1)}\,dx \)
- \( \int_0^1\frac{dx}{(x+1)(x+2)} \)
- \( \int\frac{x + 1}{x^2(x-1)}\,dx \)
- \( \int\frac{dx}{x^2 + 2x + 5} \)
Answers: (1) \( \frac12\ln\left|\frac{x-2}{x}\right| + C \); (2) \( 2\ln|x-1| + 3\ln|x+1| + C \); (3) \( \ln\frac43 \); (4) \( -2\ln|x| + \frac1x + 2\ln|x-1| + C \); (5) \( \frac12\arctan\frac{x+1}{2} + C \).
For (4), the form is \( \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x-1} \). Cover-up gives \( B = -1 \) and \( C = 2 \), and matching the \( x^2 \) coefficients gives \( A + C = 0 \), so \( A = -2 \).
FAQ
What if the numerator’s degree is too high?
Divide first. For example, \( \frac{x^2}{x^2 - 1} = 1 + \frac{1}{x^2 - 1} \).
Is there a faster way to find the constants?
The cover-up method handles distinct linear factors instantly. For the rest, substitute convenient \( x \)-values or compare coefficients.
How do I know if a quadratic factor is irreducible?
Compute its discriminant. If \( b^2 - 4ac < 0 \), it has no real roots and cannot be factored over the reals, so it gets a linear numerator. If the discriminant is zero or positive, factor it into linear pieces.
When should I not use partial fractions?
When a simpler method works. If the numerator is a multiple of the denominator’s derivative, a single substitution finishes the job. And if the integrand is not a rational function at all (it contains roots, logs or trig functions), partial fractions does not apply directly; try integration by parts or a substitution first.
Can I use partial fractions on a definite integral?
Yes. Decompose and integrate exactly as for an indefinite integral, then evaluate the antiderivative at the two limits. Combining the logarithms at the end, as in \( \ln\frac43 \), usually gives the neatest answer. Just make sure the interval does not contain a root of the denominator; if it does, the integral is improper and needs a limit.
Further reading
- Partial Fractions (Paul’s Online Math Notes) — a table of decomposition forms and six fully worked integrals.
- Integration Strategy (Paul’s Online Math Notes) — where partial fractions fits among the other integration techniques.
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