What Is Integration by Partial Fractions?
Integration by partial fractions is a technique for integrating a rational function by first splitting it into a sum of simpler fractions. A rational function is a ratio of polynomials $\frac{P(x)}{Q(x)}$, and it is rational functions of exactly this shape that the method targets. Once the fraction is written as a sum of small pieces, each piece matches a standard integral, so a hard problem becomes a few easy ones.
The method has one strict entry rule and one main idea.
Entry rule (proper first): the fraction must be proper, meaning the degree of the numerator $P$ is strictly less than the degree of the denominator $Q$. If it is improper ($\deg P \ge \deg Q$), do polynomial long division first and decompose only the proper remainder.
Main idea (partial fraction decomposition): factor $Q(x)$ completely, write one template term for each factor, and solve for the unknown constants. This is the reverse of adding fractions over a common denominator.
It sits inside the wider family of methods of integration, and it is the standard tool for integration of rational functions whenever the denominator can be factored.
How Do You Decompose A Rational Function Into Partial Fractions?
The shape of the decomposition depends entirely on how the denominator factors. There are three cases you will meet again and again.
Case 1: distinct linear factors. If $Q(x)$ is a product of different linear factors, each factor contributes one constant over that factor:
$$\frac{P(x)}{(x-a)(x-b)} = \frac{A}{x-a} + \frac{B}{x-b}$$
Case 2: repeated linear factors. A factor raised to a power needs one term for every power up to that exponent:
$$\frac{P(x)}{(x-a)^2} = \frac{A}{x-a} + \frac{B}{(x-a)^2}$$
Case 3: an irreducible quadratic factor. A quadratic with no real roots (its discriminant is negative) gets a linear numerator, not just a constant:
$$\frac{P(x)}{x^2+bx+c} = \frac{Ax+B}{x^2+bx+c}$$
To find the constants you have two reliable options. The cover-up method works fastest for distinct linear factors: to get the constant over $(x-a)$, cover that factor in the original fraction and evaluate what is left at $x=a$. The equating-coefficients method works for every case: clear the denominators, then match the coefficients of each power of $x$ on both sides to get a small system of equations.
What Are The Forms Used In Partial Fraction Decomposition?
Before any integrating, match the denominator to the right template. This one table covers the cases you need.
Table: Partial fraction template for each type of denominator factor.
Factor in the denominator | Terms it contributes to the decomposition |
|---|---|
Distinct linear $(x-a)$ | $\dfrac{A}{x-a}$ |
Repeated linear $(x-a)^k$ | $\dfrac{A_1}{x-a} + \dfrac{A_2}{(x-a)^2} + \cdots + \dfrac{A_k}{(x-a)^k}$ |
Irreducible quadratic $(x^2+bx+c)$ | $\dfrac{Ax+B}{x^2+bx+c}$ |
Repeated irreducible quadratic $(x^2+bx+c)^k$ | $\dfrac{A_1x+B_1}{x^2+bx+c} + \cdots + \dfrac{A_kx+B_k}{(x^2+bx+c)^k}$ |
Two integrals do the final work in almost every problem. A linear factor integrates to a logarithm, $\int \frac{1}{x-a},dx = \ln\lvert x-a\rvert + C$, and an irreducible quadratic integrates to an arctangent, $\int \frac{1}{x^2+p^2},dx = \frac{1}{p}\arctan\frac{x}{p} + C$. Every partial fraction integral is a combination of these two shapes, sometimes reached through integration by substitution for the quadratic pieces.
How Do You Integrate Using Partial Fractions? Worked Examples
Each example is fully stepped, and every result is checked by differentiating the antiderivative back to the original integrand. The derivative operator $\frac{d}{dx}$ is used throughout for those checks.
Example 1: Distinct linear factors.
Evaluate $\displaystyle\int \frac{1}{(x-1)(x+2)},dx$.
Set up the decomposition and clear denominators:
$$\frac{1}{(x-1)(x+2)} = \frac{A}{x-1} + \frac{B}{x+2} \quad\Longrightarrow\quad 1 = A(x+2) + B(x-1)$$
Use the cover-up values. At $x=1$: $1 = 3A$, so $A = \tfrac{1}{3}$. At $x=-2$: $1 = -3B$, so $B = -\tfrac{1}{3}$. Now integrate each piece:
$$\int \frac{1}{(x-1)(x+2)},dx = \frac{1}{3}\int \frac{dx}{x-1} - \frac{1}{3}\int \frac{dx}{x+2} = \frac{1}{3}\ln\lvert x-1\rvert - \frac{1}{3}\ln\lvert x+2\rvert + C$$
Combine the logs:
$$= \frac{1}{3}\ln\left\lvert \frac{x-1}{x+2}\right\rvert + C$$
Check: $\frac{d}{dx}\left[\frac{1}{3}\big(\ln\lvert x-1\rvert - \ln\lvert x+2\rvert\big)\right] = \frac{1}{3}\left(\frac{1}{x-1} - \frac{1}{x+2}\right) = \frac{1}{3}\cdot\frac{3}{(x-1)(x+2)} = \frac{1}{(x-1)(x+2)}$, the original integrand.
Final answer: $\displaystyle\int \frac{1}{(x-1)(x+2)},dx = \frac{1}{3}\ln\left\lvert \frac{x-1}{x+2}\right\rvert + C$.
Example 2: A numerator over distinct linear factors.
Evaluate $\displaystyle\int \frac{3x+1}{x^2+x},dx$.
Factor the denominator first: $x^2 + x = x(x+1)$. Then decompose:
$$\frac{3x+1}{x(x+1)} = \frac{A}{x} + \frac{B}{x+1} \quad\Longrightarrow\quad 3x+1 = A(x+1) + Bx$$
At $x=0$: $1 = A$. At $x=-1$: $-2 = -B$, so $B = 2$. Integrate:
$$\int \frac{3x+1}{x^2+x},dx = \int \frac{dx}{x} + 2\int \frac{dx}{x+1} = \ln\lvert x\rvert + 2\ln\lvert x+1\rvert + C$$
Check: $\frac{d}{dx}\big[\ln\lvert x\rvert + 2\ln\lvert x+1\rvert\big] = \frac{1}{x} + \frac{2}{x+1} = \frac{(x+1)+2x}{x(x+1)} = \frac{3x+1}{x^2+x}$, the original integrand.
Final answer: $\displaystyle\int \frac{3x+1}{x^2+x},dx = \ln\lvert x\rvert + 2\ln\lvert x+1\rvert + C$.
Example 3: A repeated linear factor.
Evaluate $\displaystyle\int \frac{x}{(x-1)^2},dx$.
A repeated factor needs both powers in the template:
$$\frac{x}{(x-1)^2} = \frac{A}{x-1} + \frac{B}{(x-1)^2} \quad\Longrightarrow\quad x = A(x-1) + B$$
At $x=1$: $1 = B$. Matching the coefficient of $x$ gives $A = 1$. Integrate each term, using $\int (x-1)^{-2},dx = -(x-1)^{-1}$:
$$\int \frac{x}{(x-1)^2},dx = \int \frac{dx}{x-1} + \int \frac{dx}{(x-1)^2} = \ln\lvert x-1\rvert - \frac{1}{x-1} + C$$
Check: $\frac{d}{dx}\left[\ln\lvert x-1\rvert - \frac{1}{x-1}\right] = \frac{1}{x-1} + \frac{1}{(x-1)^2} = \frac{(x-1)+1}{(x-1)^2} = \frac{x}{(x-1)^2}$, the original integrand.
Final answer: $\displaystyle\int \frac{x}{(x-1)^2},dx = \ln\lvert x-1\rvert - \frac{1}{x-1} + C$.
Example 4: An improper fraction (divide first).
Evaluate $\displaystyle\int \frac{x^2}{x^2-1},dx$.
Here $\deg P = \deg Q = 2$, so the fraction is improper. Long division comes first:
$$\frac{x^2}{x^2-1} = 1 + \frac{1}{x^2-1} = 1 + \frac{1}{(x-1)(x+1)}$$
Decompose the proper remainder: $\frac{1}{(x-1)(x+1)} = \frac{1/2}{x-1} - \frac{1/2}{x+1}$. Now integrate the whole thing:
$$\int \frac{x^2}{x^2-1},dx = \int 1,dx + \frac{1}{2}\int \frac{dx}{x-1} - \frac{1}{2}\int \frac{dx}{x+1} = x + \frac{1}{2}\ln\left\lvert \frac{x-1}{x+1}\right\rvert + C$$
Check: $\frac{d}{dx}\left[x + \frac{1}{2}\ln\left\lvert \frac{x-1}{x+1}\right\rvert\right] = 1 + \frac{1}{2}\left(\frac{1}{x-1} - \frac{1}{x+1}\right) = 1 + \frac{1}{x^2-1} = \frac{x^2-1+1}{x^2-1} = \frac{x^2}{x^2-1}$, the original integrand.
Final answer: $\displaystyle\int \frac{x^2}{x^2-1},dx = x + \frac{1}{2}\ln\left\lvert \frac{x-1}{x+1}\right\rvert + C$.
For the irreducible-quadratic case, split the term into a log part and an arctangent part. For instance, $\int \frac{2x+3}{x^2+4},dx = \ln(x^2+4) + \frac{3}{2}\arctan\frac{x}{2} + C$, since $\frac{d}{dx}\ln(x^2+4) = \frac{2x}{x^2+4}$ and $\frac{d}{dx}\left[\frac{3}{2}\arctan\frac{x}{2}\right] = \frac{3}{x^2+4}$.
Why Does Integration by Partial Fractions Work?
The method is nothing more than the addition of fractions run in reverse, used to turn one integral you cannot do into several you can.
It undoes a common denominator. When you add $\frac{A}{x-a} + \frac{B}{x-b}$ you combine them over $(x-a)(x-b)$. Partial fractions starts from the combined fraction and recovers the simple pieces, because those pieces are the ones with known integrals.
Every proper rational function has an elementary integral. Factor any real polynomial denominator and you get only linear and irreducible-quadratic factors. Linear factors integrate to logarithms; quadratic factors integrate to logarithms and arctangents. So a proper rational function always integrates in terms of $\ln$ and $\arctan$, which is exactly why the method never gets stuck.
Long division fixes the degree. An improper fraction hides a polynomial part. Dividing it out exposes that polynomial (which integrates by the power rule) and leaves a proper remainder the templates can handle. Skipping this step is why an improper fraction seems to resist decomposition.
Seen this way, integration by partial fractions is a bookkeeping move: it reorganises a rational function into the few shapes whose antiderivatives you already know.
Who Discovered Integration by Partial Fractions?
The technique grew out of the first decades of calculus, when the new subject needed a systematic way to integrate any ratio of polynomials.
Two figures anchor the history:
Johann Bernoulli (1667–1748, Switzerland) formalised partial fractions as the route to integrating rational functions, part of his wide influence on early calculus.
Gottfried Wilhelm Leibniz (1646–1716, Germany), co-inventor of calculus and the source of the $\int$ sign, worked with Bernoulli on the result that all rational functions have elementary integrals.
Where Is Integration by Partial Fractions Used In The Real World?
The method matters wherever a quantity is modelled as a ratio of polynomials, which happens across science and engineering.
Control systems and signal processing: inverting a Laplace transform almost always means decomposing a rational function of $s$ into partial fractions, then reading off the time-domain response term by term.
Electrical engineering: the current or voltage in a circuit with resistors, capacitors, and inductors is a rational function in the frequency domain, and partial fractions turn it back into real signals.
Chemical kinetics and biology: rate equations for reactions and for population or drug-clearance models produce rational integrands that partial fractions integrate exactly.
Probability and statistics: integrating certain density and generating functions relies on breaking a rational expression into pieces with known integrals.
Economics: models with rational cost or growth rates use the same decomposition to recover totals from rates.
One algebraic trick, splitting a fraction, quietly powers the mathematics behind circuits, controllers, and reaction models alike.
What Are The Most Common Mistakes With Integration by Partial Fractions?
These three errors account for most lost marks, and each matches a question real students ask on r/calculus, r/learnmath, and course common-error handouts.
Skipping long division on an improper fraction.
Where it slips in:
A student sees $\frac{x^2}{x^2-1}$ and jumps straight to $\frac{A}{x-1} + \frac{B}{x+1}$, or starts hunting for an antiderivative before decomposing at all.
Don't do this:
Do not decompose an improper fraction. The templates only produce a proper fraction, so they can never reconstruct one whose numerator degree is as large as the denominator's.
The correct way:
Compare degrees first. If $\deg P \ge \deg Q$, divide to get a polynomial plus a proper remainder, then decompose only the remainder: $\frac{x^2}{x^2-1} = 1 + \frac{1}{(x-1)(x+1)}$.
Using the wrong form for a repeated or quadratic factor.
Where it slips in:
A student writes only $\frac{B}{(x-1)^2}$ for a repeated factor and drops the $\frac{A}{x-1}$ term, or writes $\frac{A}{x^2+4}$ for an irreducible quadratic instead of $\frac{Ax+B}{x^2+4}$.
Don't do this:
Do not give a repeated factor a single term, and do not give a quadratic factor a constant numerator. Both leave the system with too few unknowns to ever match.
The correct way:
Include every power for a repeat, $\frac{A}{x-1} + \frac{B}{(x-1)^2}$, and give each irreducible quadratic a full linear numerator, $\frac{Ax+B}{x^2+bx+c}$.
Sign errors while solving for the constants.
Where it slips in:
Clearing denominators such as $1 = A(x+2) + B(x-1)$ and then mishandling a negative substitution, for example writing $1 = 3B$ at $x=-2$ instead of $1 = -3B$.
Don't do this:
Do not rush the substitution step. A single dropped minus sign flips a constant and corrupts the whole integral.
The correct way:
Substitute the root of each factor carefully and, as a safety check, add your partial fractions back over the common denominator to confirm the numerator returns to the original $P(x)$.
Practice Problems On Integration by Partial Fractions
Work each one, then check against the answer. Answers are verified by differentiating back.
$\displaystyle\int \frac{1}{(x-2)(x+1)},dx$.
(Answer: $\frac{1}{3}\ln\left\lvert\frac{x-2}{x+1}\right\rvert + C$.)$\displaystyle\int \frac{1}{x^2-9},dx$.
(Answer: $\frac{1}{6}\ln\left\lvert\frac{x-3}{x+3}\right\rvert + C$.)$\displaystyle\int \frac{5x-1}{x^2-x-2},dx$.
(Answer: factor to $(x-2)(x+1)$; $3\ln\lvert x-2\rvert + 2\ln\lvert x+1\rvert + C$.)$\displaystyle\int \frac{1}{x(x-1)^2},dx$.
(Answer: $\ln\lvert x\rvert - \ln\lvert x-1\rvert - \frac{1}{x-1} + C$.)$\displaystyle\int \frac{x^2+1}{x^2-1},dx$.
(Answer: improper; $x + \ln\left\lvert\frac{x-1}{x+1}\right\rvert + C$.)$\displaystyle\int \frac{1}{x^2+2x+5},dx$.
(Answer: complete the square to $(x+1)^2+4$; $\frac{1}{2}\arctan\frac{x+1}{2} + C$.)
Where Should You Go Next After Integration by Partial Fractions?
Partial fractions is one tool in a larger toolkit, and several natural doors open from here.
Methods of integration. See how partial fractions sits beside the other main techniques and when to reach for each.
Integration by parts. The go-to method for products such as $x e^x$ or $x\ln x$, which partial fractions cannot touch.
Integration of rational functions. Go deeper on the full rational-function workflow, including trickier repeated and quadratic denominators.
Indefinite integrals. Reinforce the $+C$ and the antiderivative ideas every partial fraction result relies on.
If you are meeting integration by partial fractions for the first time, a live Bhanzu trainer teaches it from the decomposition idea up, so the templates feel like common sense, in the Bhanzu math program.
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