What Is Integration Of Rational Functions?
Integration of rational functions is the process of finding the integral of a rational function: a quotient $\frac{P(x)}{Q(x)}$ where $P$ and $Q$ are polynomials and $Q$ is not the zero polynomial. The strategy never changes, so it is one of the most reliable techniques in the whole of methods of integration: rewrite the awkward fraction as a sum of standard pieces, then integrate each piece from a short table.
The full workflow has five steps:
Check proper or improper. If the numerator's degree is at least the denominator's degree ($\deg P \ge \deg Q$), do polynomial long division first.
Factor the denominator $Q(x)$ into linear factors and irreducible quadratic factors.
Decompose the proper fraction into partial fractions, one simple term per factor.
Integrate each partial fraction. Linear factors give logarithms; irreducible quadratics give a logarithm plus an arctangent.
Verify by differentiating the antiderivative back to the original integrand.
Step 3 is a topic in its own right. The mechanics of finding the constants are covered in depth on the integration by partial fractions page; this article is the wider workflow that surrounds it, from the first proper-or-improper test to the final logarithm-and-arctangent answer. For the objects being integrated, see the overview of rational functions.
Because a definite integral is the signed area under a curve, integrating $\frac{P(x)}{Q(x)}$ measures the area beneath a rational graph. The power of the method is that this area is always the sum of a few simple areas: the area under $\frac{1}{x-a}$ (a logarithm) and the area under $\frac{1}{x^2+a^2}$ (an arctangent).
How Do You Know When To Use Long Division First?
Partial fraction decomposition only works on a proper rational function, one where $\deg P < \deg Q$. If the fraction is improper ($\deg P \ge \deg Q$), you must divide first.
Polynomial long division of polynomials rewrites any improper fraction as
$$\frac{P(x)}{Q(x)} = S(x) + \frac{R(x)}{Q(x)}, \qquad \deg R < \deg Q,$$
where $S(x)$ is a polynomial (easy to integrate with the power rule) and $\frac{R(x)}{Q(x)}$ is now a proper fraction ready for partial fractions.
Example 1: An improper fraction needing long division.
Evaluate $\displaystyle\int \frac{x^2 + 3x + 5}{x + 1},dx$.
The numerator has degree 2 and the denominator degree 1, so the fraction is improper. Divide $x^2 + 3x + 5$ by $x + 1$:
$$\frac{x^2 + 3x + 5}{x + 1} = x + 2 + \frac{3}{x + 1}$$
Now integrate term by term:
$$\int \left(x + 2 + \frac{3}{x + 1}\right) dx = \frac{x^2}{2} + 2x + 3\ln\lvert x + 1\rvert + C$$
Check by differentiating back: $F'(x) = x + 2 + \dfrac{3}{x+1} = \dfrac{(x+2)(x+1) + 3}{x+1} = \dfrac{x^2 + 3x + 5}{x+1}$, the original integrand.
Final answer: $\displaystyle\int \frac{x^2 + 3x + 5}{x + 1},dx = \frac{x^2}{2} + 2x + 3\ln\lvert x + 1\rvert + C$.
Skip this division step on an improper fraction and every later line is wrong, which is why it heads the mistakes section below.
How Do You Decompose Into Partial Fractions?
Once the fraction is proper, factor $Q(x)$ completely and write one group of terms for each factor. The form of the term depends on the factor.
Table: The partial fraction setup for each type of denominator factor.
Factor In $Q(x)$ | Terms In 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 $(x^2 + bx + c)$ | $\dfrac{Bx + C}{x^2 + bx + c}$ |
Repeated irreducible quadratic $(x^2 + bx + c)^k$ | $\dfrac{B_1x + C_1}{x^2 + bx + c} + \cdots + \dfrac{B_kx + C_k}{(x^2 + bx + c)^k}$ |
A quadratic $x^2 + bx + c$ is irreducible over the real numbers when its discriminant is negative, $b^2 - 4ac < 0$, so it has no real roots and cannot be split into real linear factors. Recognising that condition is a small application of the theory behind quadratic equations. Note the linear numerator $Bx + C$ over a quadratic factor: a bare constant is not enough.
To find the constants, multiply both sides by $Q(x)$ and either substitute the roots (the cover-up shortcut) or compare coefficients. The partial fractions algebra reference works these solves case by case.
Example 2: Distinct linear factors.
Evaluate $\displaystyle\int \frac{3x + 2}{x^3 - x^2 - 2x},dx$.
Factor the denominator: $x^3 - x^2 - 2x = x(x^2 - x - 2) = x(x - 2)(x + 1)$. Set up the decomposition:
$$\frac{3x + 2}{x(x - 2)(x + 1)} = \frac{A}{x} + \frac{B}{x - 2} + \frac{C}{x + 1}$$
Multiply through by $x(x-2)(x+1)$:
$$3x + 2 = A(x - 2)(x + 1) + Bx(x + 1) + Cx(x - 2)$$
Substitute the roots one at a time:
$x = 0$: $;2 = A(-2)(1) \Rightarrow A = -1$.
$x = 2$: $;8 = B(2)(3) \Rightarrow B = \tfrac{4}{3}$.
$x = -1$: $;-1 = C(-1)(-3) \Rightarrow C = -\tfrac{1}{3}$.
Integrate each simple fraction:
$$\int \frac{3x+2}{x^3 - x^2 - 2x},dx = -\ln\lvert x\rvert + \tfrac{4}{3}\ln\lvert x - 2\rvert - \tfrac{1}{3}\ln\lvert x + 1\rvert + C$$
Check: differentiating gives $-\dfrac{1}{x} + \dfrac{4/3}{x-2} - \dfrac{1/3}{x+1}$, which recombines over $x(x-2)(x+1)$ to $\dfrac{3x+2}{x(x-2)(x+1)}$.
Final answer: $-\ln\lvert x\rvert + \tfrac{4}{3}\ln\lvert x - 2\rvert - \tfrac{1}{3}\ln\lvert x + 1\rvert + C$.
Example 3: A repeated linear factor.
Evaluate $\displaystyle\int \frac{3x + 5}{(x - 1)^2},dx$.
A squared factor needs two terms, one for each power:
$$\frac{3x + 5}{(x - 1)^2} = \frac{A}{x - 1} + \frac{B}{(x - 1)^2}$$
Multiply through: $3x + 5 = A(x - 1) + B$. At $x = 1$, $;8 = B$. Comparing the $x$ coefficient gives $A = 3$. So
$$\int \frac{3x+5}{(x-1)^2},dx = \int \left(\frac{3}{x-1} + \frac{8}{(x-1)^2}\right) dx = 3\ln\lvert x - 1\rvert - \frac{8}{x - 1} + C$$
Check: $F'(x) = \dfrac{3}{x-1} + \dfrac{8}{(x-1)^2} = \dfrac{3(x-1) + 8}{(x-1)^2} = \dfrac{3x + 5}{(x-1)^2}$.
Final answer: $3\ln\lvert x - 1\rvert - \dfrac{8}{x - 1} + C$.
How Do You Integrate Each Partial Fraction?
After the split, every piece matches one row of a short table. These are the only antiderivatives the method ever needs, and each carries $+C$ as an indefinite integral.
Table: The standard integrals every partial fraction reduces to.
Partial Fraction Piece | Antiderivative |
|---|---|
$\dfrac{A}{x - a}$ | $A\ln\lvert x - a\rvert + C$ |
$\dfrac{A}{(x - a)^n},; n \ge 2$ | $\dfrac{-A}{(n-1)(x-a)^{n-1}} + C$ |
$\dfrac{Bx}{x^2 + a^2}$ | $\dfrac{B}{2}\ln(x^2 + a^2) + C$ |
$\dfrac{C}{x^2 + a^2}$ | $\dfrac{C}{a}\arctan\left(\dfrac{x}{a}\right) + C$ |
The logarithm rows are why so much of this topic reduces to properties of logarithms. The arctangent row is the one students most often forget, because a quadratic denominator quietly produces an inverse-trig answer rather than another log. For the full reference set, see the list of integrals.
How Do You Handle An Irreducible Quadratic Denominator?
An irreducible quadratic factor produces a numerator $Bx + C$. Split that single fraction into two pieces: a $Bx$ piece that integrates to a logarithm, and a constant piece that integrates to an arctangent.
Example 4: A linear factor and an irreducible quadratic together.
Evaluate $\displaystyle\int \frac{2x - 3}{x^3 + x},dx$.
Factor: $x^3 + x = x(x^2 + 1)$, and $x^2 + 1$ is irreducible (discriminant $-4 < 0$). Decompose with a constant over the linear factor and a linear numerator over the quadratic:
$$\frac{2x - 3}{x(x^2 + 1)} = \frac{A}{x} + \frac{Bx + C}{x^2 + 1}$$
Multiply through: $2x - 3 = A(x^2 + 1) + (Bx + C)x = (A + B)x^2 + Cx + A$. Matching coefficients:
constant: $A = -3$,
$x$: $C = 2$,
$x^2$: $A + B = 0 \Rightarrow B = 3$.
Now integrate, splitting the quadratic piece into its logarithm half and arctangent half:
$$\int \frac{2x-3}{x^3 + x},dx = \int\left(\frac{-3}{x} + \frac{3x}{x^2+1} + \frac{2}{x^2+1}\right) dx = -3\ln\lvert x\rvert + \frac{3}{2}\ln(x^2 + 1) + 2\arctan x + C$$
Check: $F'(x) = -\dfrac{3}{x} + \dfrac{3x}{x^2+1} + \dfrac{2}{x^2+1} = -\dfrac{3}{x} + \dfrac{3x + 2}{x^2+1}$. Over the common denominator $x(x^2+1)$ this is $\dfrac{-3(x^2+1) + (3x+2)x}{x(x^2+1)} = \dfrac{2x - 3}{x(x^2+1)}$.
Final answer: $-3\ln\lvert x\rvert + \dfrac{3}{2}\ln(x^2 + 1) + 2\arctan x + C$.
When the quadratic is not already in the clean $x^2 + a^2$ form, complete the square and use a $u$-substitution. For instance, $x^2 + 6x + 13 = (x + 3)^2 + 4$, so with $u = x + 3$:
$$\int \frac{dx}{x^2 + 6x + 13} = \int \frac{du}{u^2 + 4} = \frac{1}{2}\arctan\left(\frac{x + 3}{2}\right) + C$$
Check: $F'(x) = \dfrac{1}{2}\cdot\dfrac{1/2}{1 + \left(\frac{x+3}{2}\right)^2} = \dfrac{1}{(x+3)^2 + 4} = \dfrac{1}{x^2 + 6x + 13}$. The substitution mechanics are covered under integration by substitution.
Why Does Integration Of Rational Functions Always Work?
Unlike many integrals, a rational function is guaranteed to have an elementary antiderivative. Two facts make the method total.
Every real polynomial factors into linear and irreducible quadratic pieces. This is a consequence of the fundamental theorem of algebra. So the denominator can always be reduced to exactly the factor types in the decomposition table, with nothing left over.
Each of those pieces has a known antiderivative. A linear factor integrates to a logarithm, a repeated linear factor to a negative power, and an irreducible quadratic to a logarithm plus an arctangent. There is no fifth kind of piece to worry about.
The geometry sits underneath the algebra. The area under $\dfrac{1}{x - a}$ grows like a logarithm, because the strips get thinner in exactly the proportion that builds $\ln\lvert x - a\rvert$. The area under $\dfrac{1}{x^2 + a^2}$ sweeps out an angle, which is why an arctangent appears. Partial fractions is the act of cutting one complicated area into these two familiar shapes.
That completeness is also why the topic is worth mastering early: once a fraction of polynomials appears in a problem, you always know the integral exists and how to reach it.
Who Invented Integration By Partial Fractions?
The method is older than most calculus students expect, and it grew directly out of the first decades of the subject.
Two other names anchor the method:
Gottfried Wilhelm Leibniz (1646–1716, Germany) supplied both the notation ($\int$ and $dx$) and, alongside Bernoulli, the early treatment of fractional integrands.
Carl Friedrich Gauss (1777–1855, Germany) proved the fundamental theorem of algebra in 1799, which guarantees the linear-and-irreducible-quadratic factorisation the whole method depends on.
Where Is Integration Of Rational Functions Used In The Real World?
Ratios of polynomials appear wherever one changing quantity is governed by another, so the integral shows up across the applied sciences.
Control and electrical engineering: transfer functions are rational, and inverting them (via partial fractions) is the standard way to find how a circuit or system responds over time.
Chemical kinetics: reaction-rate laws in which the rate depends on a ratio of concentrations integrate to logarithms of those concentrations, giving the time a reaction takes.
Population and epidemic models: the logistic growth equation separates into a rational integrand, and partial fractions is exactly the step that produces its S-shaped solution.
Probability: several continuous densities are rational functions, and their cumulative probabilities are arctangent and logarithm integrals of the kind above.
Physics of motion with resistance: a body slowed by a drag force proportional to a polynomial in speed leads to a rational integral for the time or distance travelled.
One procedure, taught in a single chapter, quietly underwrites circuit design, reaction timing, and population forecasting.
What Are The Most Common Mistakes With Integration Of Rational Functions?
These four errors account for most lost marks, and each matches a warning that recurs across course handouts and student forums.
Skipping long division on an improper fraction.
Where it slips in:
A student sees $\dfrac{x^2 + 3x + 5}{x + 1}$ and jumps straight to partial fractions, even though the numerator's degree is not smaller than the denominator's.
Don't do this:
Do not decompose an improper fraction. Partial fractions is only valid when $\deg P < \deg Q$.
The correct way:
Divide first: $\dfrac{x^2 + 3x + 5}{x + 1} = x + 2 + \dfrac{3}{x + 1}$, then integrate the polynomial part and the proper remainder.
Putting a constant over an irreducible quadratic.
Where it slips in:
A student writes $\dfrac{A}{x^2 + 1}$ for the factor $x^2 + 1$, using a single constant on top as if it were a linear factor.
Don't do this:
Do not use a bare constant over a quadratic factor. The numerator has one degree less than the denominator, so it must be linear.
The correct way:
Use $\dfrac{Bx + C}{x^2 + 1}$, then split it into a $\dfrac{Bx}{x^2+1}$ log piece and a $\dfrac{C}{x^2+1}$ arctangent piece.
Forgetting the extra terms for a repeated factor.
Where it slips in:
For $(x - 1)^2$ in the denominator, a student writes only $\dfrac{A}{x - 1}$ and never accounts for the squared power.
Don't do this:
Do not represent a repeated factor with a single term. One term cannot reproduce every proper numerator over $(x-1)^2$.
The correct way:
Include one term per power: $\dfrac{A}{x - 1} + \dfrac{B}{(x - 1)^2}$, up to the multiplicity of the factor.
Dropping the arctangent (or the absolute value in the log).
Where it slips in:
A student integrates $\dfrac{Bx + C}{x^2 + a^2}$, handles the $\ln$ half, and forgets that the constant half becomes an arctangent, or writes $\ln(x - a)$ without the absolute value.
Don't do this:
Do not stop after the logarithm, and do not omit $\lvert;\rvert$ around a linear argument.
The correct way:
Split the quadratic piece fully, $\dfrac{B}{2}\ln(x^2 + a^2) + \dfrac{C}{a}\arctan\left(\dfrac{x}{a}\right)$, and keep $\ln\lvert x - a\rvert$ for linear factors.
Practice Problems On Integration Of Rational Functions
Work each one, then check against the verified answer.
$\displaystyle\int \frac{dx}{(x - 1)(x + 2)}$.
(Answer: $\tfrac{1}{3}\ln\left\lvert \dfrac{x - 1}{x + 2}\right\rvert + C$.)$\displaystyle\int \frac{5x - 4}{x^2 - x - 2},dx$.
(Answer: $2\ln\lvert x - 2\rvert + 3\ln\lvert x + 1\rvert + C$.)$\displaystyle\int \frac{x^2}{x - 1},dx$ (improper).
(Answer: $\dfrac{x^2}{2} + x + \ln\lvert x - 1\rvert + C$.)$\displaystyle\int \frac{4}{x^2 + 4},dx$.
(Answer: $2\arctan\left(\dfrac{x}{2}\right) + C$.)$\displaystyle\int \frac{2x + 1}{x^2 + 1},dx$.
(Answer: $\ln(x^2 + 1) + \arctan x + C$.)$\displaystyle\int \frac{dx}{x^2 + 6x + 13}$.
(Answer: $\dfrac{1}{2}\arctan\left(\dfrac{x + 3}{2}\right) + C$.)
Where Should You Go Next After Integration Of Rational Functions?
This method is a hub, and several natural doors open from here.
Integration by partial fractions. Go deeper on the decomposition step, including the cover-up shortcut and repeated-factor solves.
Methods of integration. Place this technique beside substitution, parts, and trigonometric methods in one map.
Integration by parts. The other main technique for products, useful when a rational function multiplies a logarithm or an exponential.
Definite integrals. Apply the antiderivatives here to compute exact areas between limits.
If you are meeting integration of rational functions for the first time, a live Bhanzu trainer teaches it from the workflow up, so each step feels inevitable rather than memorised, in the Bhanzu math program.
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