List of Integrals of Rational Functions With Proofs

What Is The List Of Integrals Of Rational Functions?

The list of integrals of rational functions is a reference table of antiderivatives for integrands of the form $\frac{P(x)}{Q(x)}$, where $P$ and $Q$ are polynomials and $Q$ is not the zero polynomial. Such a quotient is a rational function. The anchor entry is

$$\int \frac{dx}{ax+b} = \frac{1}{a}\ln\lvert ax+b\rvert + C, \qquad a \neq 0,$$

and the second anchor is $\int \frac{dx}{x^2+a^2} = \frac{1}{a}\arctan\frac{x}{a} + C$. Every other row is one of these two, a power-rule row, or a combination of them reached through algebra.

The symbols used in every table:

This page is the catalogue. For the full lesson on the technique, see integration of rational functions.

How Do You Choose The Right Row For A Rational Integral?

Run three checks in order, and the row picks itself.

1. Compare degrees. If the degree of $P$ is at least the degree of $Q$, the fraction is improper. Divide first with long division of polynomials, integrate the polynomial part, and keep only the proper remainder.

2. Factor the denominator completely into linear factors and irreducible quadratics (quadratics with $D < 0$).

3. Match each piece. A single factor matches a table row directly; a product of factors goes through partial fractions first.

Two reading rules sit on top. The logarithm rows keep the absolute value, because $\frac{1}{x}$ is defined on both sides of $0$. And answers that look different can both be right, since $\ln\left\lvert\frac{x-a}{x+a}\right\rvert$ and $-\ln\left\lvert\frac{x+a}{x-a}\right\rvert$ are the same function.

What Are The Integrals With Linear Denominators?

A linear denominator gives a logarithm, scaled by $\frac{1}{a}$ for the chain rule. The rows come from integration by substitution with $u = ax + b$, sometimes after splitting the numerator.

The last three rows are partial fractions already done for you. They save time on homework, and they are also the pattern the general method follows.

What Are The Integrals Of Powers Of $(ax+b)$?

Powers of a linear factor follow the power rule after $u = ax + b$, with one exception: the power $-1$, which is the logarithm row above.

The $\frac{x}{(ax+b)^2}$ row is worth deriving once, because it shows how a numerator is rewritten in terms of the denominator. Write $x = \frac{1}{a}(ax+b) - \frac{b}{a}$, split the fraction into $\frac{1}{a(ax+b)} - \frac{b}{a(ax+b)^2}$, and each piece is already in the table.

What Are The Integrals Involving $x^2 + a^2$?

An irreducible $x^2 + a^2$ produces an arctangent when the numerator is a constant and a logarithm when the numerator is a multiple of $x$. The arctan row is the derivative of the inverse tangent read backwards: $\frac{d}{dx}\arctan\frac{x}{a} = \frac{a}{x^2+a^2}$.

A Physics Forums student put the confusion in this family plainly: "I'm completely confused as to how u/u²+1 becomes the log function?" The answer is the numerator. In $\frac{u}{u^2+1}$, the numerator $u$ is half the derivative of $u^2 + 1$, so the substitution $z = u^2 + 1$ turns it into $\frac{1}{2}\int\frac{dz}{z}$, a logarithm. In $\frac{1}{u^2+1}$ there is no such factor, and the result is $\arctan u$.

Higher powers use the reduction formula, which lowers the power by one each time:

$$\int \frac{dx}{(x^2+a^2)^{n}} = \frac{x}{2a^2(n-1)(x^2+a^2)^{n-1}} + \frac{2n-3}{2a^2(n-1)}\int \frac{dx}{(x^2+a^2)^{n-1}}, \qquad n \geq 2.$$

It comes from integration by parts on the $(n-1)$ case; trigonometric substitution with $x = a\tan\theta$ reaches the same results.

What Are The Integrals Involving $x^2 - a^2$?

$x^2 - a^2$ factors as $(x-a)(x+a)$, so its rows are logarithms of a ratio rather than arctangents. A Collegedunia question asks to "Evaluate the integral $\int dx/(a^2 - x^2)$"; partial fractions give $\frac{1}{a^2-x^2} = \frac{1}{2a}\left(\frac{1}{a-x} + \frac{1}{a+x}\right)$, which integrates to the second row below.

For $\lvert x\rvert < a$ the second row also equals $\frac{1}{a}\operatorname{artanh}\frac{x}{a}$, the inverse hyperbolic function form some textbooks print instead. The two are the same function written two ways.

What Are The Integrals With A General Quadratic Denominator?

For $ax^2 + bx + c$, the discriminant $D = b^2 - 4ac$ decides the row: negative gives an arctangent, positive gives a logarithm, and zero gives a power. Each row comes from completing the square, which rewrites the quadratic as $a\left(x + \frac{b}{2a}\right)^2 - \frac{D}{4a}$.

The linear-over-quadratic row is the workhorse. Split the numerator into a multiple of the denominator's derivative, $2ax + b$, plus a constant; the first part gives a logarithm and the constant part goes back to the first three rows.

What Are The Partial Fraction Patterns For Rational Integrals?

Partial fractions rewrite a proper fraction with a factored denominator as a sum of simpler fractions, one family per factor. The Physics Forums thread "Integration of Rational Functions by Partial Fractions" shows the usual first hurdle, an improper fraction, which must be divided before any splitting (it is Example 3 below). The patterns, with the integral each term produces:

The constants come from multiplying through by $Q(x)$ and either substituting the roots (the cover-up shortcut for distinct linear factors) or matching coefficients. The algebra side is covered in partial fractions; the calculus side, with more worked cases, is integration by partial fractions.

Two cubic and quartic rows show the patterns in action:

The $x^4 + 1$ row exists only because $x^4 + 1 = (x^2 + \sqrt{2}x + 1)(x^2 - \sqrt{2}x + 1)$, the factorization Leibniz missed.

What Are The Standard Definite Integrals Of Rational Functions?

Definite rational integrals are evaluated from the rows above, with one caution: the interval must not contain a zero of the denominator, or the integral is improper and may diverge.

The first row is a small surprise worth keeping: a rational integral equals $\frac{\pi}{4}$, which is why $\arctan$ turns up in formulas for $\pi$. The infinite-limit rows are improper integrals that converge because the denominator grows at least like $x^2$.

Examples Of Integrals Of Rational Functions

Example 1

Use the list to find $\int \frac{3}{2x+5},dx$.

Match the row $\int \frac{dx}{ax+b}$ with $a = 2$ and $b = 5$:

$$\int \frac{3}{2x+5},dx = 3 \cdot \frac{1}{2}\ln\lvert 2x+5\rvert + C$$

$$= \frac{3}{2}\ln\lvert 2x+5\rvert + C$$

Final answer: $\frac{3}{2}\ln\lvert 2x+5\rvert + C$

Example 2

Find $\int \frac{dx}{x^2+4}$.

The first instinct reads "one over something" and reaches for the logarithm:

$$\text{Attempt: } \int \frac{dx}{x^2+4} = \ln\lvert x^2+4\rvert + C$$

Differentiate the attempt to check it:

$$\frac{d}{dx}\ln(x^2+4) = \frac{2x}{x^2+4}$$

That is not the integrand; the factor $2x$ appears from nowhere. The logarithm row needs the numerator to be the derivative of the denominator, and here the numerator is the constant $1$. With $x^2 + a^2$ and $a = 2$, the arctangent row applies:

$$\int \frac{dx}{x^2+4} = \frac{1}{2}\arctan\frac{x}{2} + C$$

Check: $\frac{d}{dx}\left[\frac{1}{2}\arctan\frac{x}{2}\right] = \frac{1}{2} \cdot \frac{2}{x^2+4} = \frac{1}{x^2+4}$.

Final answer: $\frac{1}{2}\arctan\frac{x}{2} + C$

Example 3

Evaluate $\int \frac{x^3+36}{x^2+36},dx$ (the problem from the Physics Forums thread).

The fraction is improper, so divide first:

$$\frac{x^3+36}{x^2+36} = x - \frac{36x - 36}{x^2+36}$$

$$= x - \frac{36x}{x^2+36} + \frac{36}{x^2+36}$$

Integrate each piece with its row ($a = 6$):

$$\int x,dx - 36 \cdot \frac{1}{2}\ln(x^2+36) + 36 \cdot \frac{1}{6}\arctan\frac{x}{6}$$

$$= \frac{x^2}{2} - 18\ln(x^2+36) + 6\arctan\frac{x}{6} + C$$

Final answer: $\frac{x^2}{2} - 18\ln(x^2+36) + 6\arctan\frac{x}{6} + C$

Example 4

Find $\int \frac{x^2+2x-1}{x^3-x},dx$.

Factor completely: $x^3 - x = x(x-1)(x+1)$. Set up distinct linear partial fractions:

$$\frac{x^2+2x-1}{x(x-1)(x+1)} = \frac{A}{x} + \frac{B}{x-1} + \frac{E}{x+1}$$

Cover-up at each root: $x = 0$ gives $A = \frac{-1}{(-1)(1)} = 1$; $x = 1$ gives $B = \frac{2}{(1)(2)} = 1$; $x = -1$ gives $E = \frac{-2}{(-1)(-2)} = -1$.

$$\int \left(\frac{1}{x} + \frac{1}{x-1} - \frac{1}{x+1}\right)dx = \ln\lvert x\rvert + \ln\lvert x-1\rvert - \ln\lvert x+1\rvert + C$$

Final answer: $\ln\left\lvert\frac{x(x-1)}{x+1}\right\rvert + C$

Example 5

Find $\int \frac{5x+1}{(x-1)^2(x+2)},dx$.

A repeated factor needs one term per power:

$$\frac{5x+1}{(x-1)^2(x+2)} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{E}{x+2}$$

Cover-up gives $B = \frac{6}{3} = 2$ at $x = 1$ and $E = \frac{-9}{9} = -1$ at $x = -2$. Comparing the $x^2$ coefficients, $A + E = 0$, so $A = 1$.

$$\int \left(\frac{1}{x-1} + \frac{2}{(x-1)^2} - \frac{1}{x+2}\right)dx = \ln\lvert x-1\rvert - \frac{2}{x-1} - \ln\lvert x+2\rvert + C$$

Final answer: $\ln\left\lvert\frac{x-1}{x+2}\right\rvert - \frac{2}{x-1} + C$

Example 6

Evaluate $\int_{0}^{1} \frac{x^2+2x+3}{(x+1)(x^2+1)},dx$.

Split with one linear and one irreducible quadratic term:

$$\frac{x^2+2x+3}{(x+1)(x^2+1)} = \frac{A}{x+1} + \frac{Bx+E}{x^2+1}$$

At $x = -1$: $A = \frac{2}{2} = 1$. Then $x^2 + 2x + 3 - (x^2 + 1) = 2x + 2$, so $(Bx + E)(x+1) = 2(x+1)$ and $B = 0$, $E = 2$.

$$\int_{0}^{1} \left(\frac{1}{x+1} + \frac{2}{x^2+1}\right)dx = \Big[\ln\lvert x+1\rvert + 2\arctan x\Big]_{0}^{1}$$

$$= \ln 2 + 2 \cdot \frac{\pi}{4} = \ln 2 + \frac{\pi}{2}$$

Final answer: $\ln 2 + \frac{\pi}{2} \approx 2.2639$

Why Does Every Rational Function Have A Formula For Its Integral?

Every rational function integrates to polynomials, logarithms and arctangents because every real polynomial factors into real linear and irreducible quadratic pieces. That is a consequence of the fundamental theorem of algebra, and it is the reason the tables on this page are complete rather than a sample.

• Factoring guarantees the split. Once $Q(x)$ is factored, partial fractions always exist, and each term is one of the four patterns above.

• Each pattern has a row. Linear terms give logarithms and powers; quadratic terms give a logarithm, an arctangent and, if repeated, the reduction formula.

• The boundary is sharp. Put a rational function under a square root and this guarantee disappears, which is why $\int \frac{dx}{\sqrt{1-x^4}}$ has no elementary formula while $\int \frac{dx}{1-x^4}$ does.

Where Do Rational Integrals Show Up In Real Life?

Rational integrals appear whenever a rate is a ratio of polynomials, which happens most often when a growth or reaction rate depends on two competing amounts.

• Population limits. The logistic model $\frac{dP}{dt} = kP\left(1 - \frac{P}{K}\right)$ is solved by integrating $\frac{1}{P(K-P)}$, a partial-fraction row from the linear table.

• Chemistry. A reaction whose rate is proportional to the product of two concentrations leads to $\int \frac{dx}{(a-x)(b-x)}$, the last row of the linear table.

• Control engineering. Engineers invert Laplace transforms of circuit and controller responses by splitting rational functions into partial fractions, term by term.

• Projectile motion with drag. A falling object with air resistance proportional to $v^2$ leads to $\int \frac{dv}{g - kv^2}$, an $a^2 - x^2$ row.

Rational Integrals: Where Things Go Sideways

Mistake 1: Writing A Logarithm For Every Reciprocal

Where it slips in: Integrals like $\frac{1}{x^2+4}$ or $\frac{1}{x^2+x+1}$, right after a page of $\frac{1}{ax+b}$ practice.

Don't do this: Writing $\int \frac{dx}{x^2+4} = \ln\lvert x^2+4\rvert + C$.

The correct way: Use the logarithm only when the numerator is the derivative of the denominator; a constant over an irreducible quadratic gives an arctangent.

The memorizer learns "one over something gives a log" as a slogan, and the slogan is exactly what fails here. The habit that fixes it is to ask what the derivative of the denominator is before writing any logarithm.

Mistake 2: Skipping Long Division

Where it slips in: Improper fractions such as $\frac{x^3+36}{x^2+36}$ or $\frac{x^2+1}{x^2-1}$.

Don't do this: Setting up $\frac{x^2+1}{x^2-1} = \frac{A}{x-1} + \frac{B}{x+1}$ straight away.

The correct way: Divide first: $\frac{x^2+1}{x^2-1} = 1 + \frac{2}{x^2-1}$, then split only the proper remainder.

The rusher sets up partial fractions at once, solves for $A$ and $B$, and ends with an answer that differentiates to the wrong function. The missing piece is always the polynomial part of the division.

Mistake 3: Splitting Before Factoring Completely

Where it slips in: Denominators like $x^3 - x$ left as $x(x^2 - 1)$, or $x^4 + 1$ assumed to be irreducible.

Don't do this: Writing $\frac{Bx+E}{x^2-1}$ as if $x^2 - 1$ were irreducible.

The correct way: Factor until every quadratic has $D < 0$: $x^2 - 1 = (x-1)(x+1)$, and even $x^4 + 1$ splits into two real quadratics.

Mistake 4: Integrating Across A Zero Of The Denominator

Where it slips in: Definite integrals such as $\int_{-1}^{1} \frac{dx}{x^2}$, where $x = 0$ sits inside the interval.

Don't do this: Computing $\left[-\frac{1}{x}\right]_{-1}^{1} = -2$, a negative area for a positive integrand.

The correct way: Split at the zero and treat each piece as an improper integral; here both pieces diverge, so the integral diverges.

Mistake 3 has a famous real-world version. In 1702, Leibniz, one of the two inventors of calculus, stated that $x^4 + a^4$ could not be factored into real linear and quadratic pieces, which would have put some rational integrals out of reach. Decades later Euler wrote down the factorization, as recorded in the history of the fundamental theorem of algebra.

Practice Problems On The List Of Integrals Of Rational Functions

1. Find $\int \frac{dx}{x^2+9}$.

2. Find $\int \frac{dx}{x^2-9}$.

3. Find $\int \frac{2x+1}{x^2+x-6},dx$.

4. Find $\int \frac{x}{(x+1)^2},dx$.

5. Find $\int \frac{dx}{x^2+2x+5}$.

6. Evaluate $\int_{0}^{1} \frac{dx}{(x+1)(x+2)}$.

Answer to Question 1: $\frac{1}{3}\arctan\frac{x}{3} + C$

Answer to Question 2: $\frac{1}{6}\ln\left\lvert\frac{x-3}{x+3}\right\rvert + C$

Answer to Question 3: $\ln\lvert x^2+x-6\rvert + C$

Answer to Question 4: $\ln\lvert x+1\rvert + \frac{1}{x+1} + C$

Answer to Question 5: $\frac{1}{2}\arctan\frac{x+1}{2} + C$

Answer to Question 6: $\ln\frac{4}{3} \approx 0.2877$

Key Takeaways

• The list of integrals of rational functions reduces to polynomials, logarithms and arctangents.

• Divide first when the fraction is improper, then factor the denominator completely.

• A numerator that is the denominator's derivative gives a logarithm; a constant over an irreducible quadratic gives an arctangent.

• The discriminant decides between the arctangent, logarithm and power rows for a general quadratic.

• Never integrate straight across a zero of the denominator.

Your Next Step With Rational Integrals

Work through practice problems 1 to 6; if a logarithm answer refuses to differentiate back, return to "What Are The Integrals Involving x^2 + a^2?" and check the numerator first. Want a live Bhanzu trainer to help you with partial fractions? Book a free demo class.

Read More

• List of integrals for every function family in one table

• Table of integrals for the standard trig, inverse-trig and quadratic forms

• Basic integration formulas for the power and reciprocal rules these rows build on

• Methods of integration for choosing between substitution, parts and partial fractions

• Indefinite integrals for the meaning of the constant C in every row