List of Integrals of Irrational Functions With Examples

What Is The List Of Integrals Of Irrational Functions?

The list of integrals of irrational functions is a reference table of antiderivatives for algebraic functions that contain a root, most often a square root of a linear or quadratic expression. Each entry pairs an integrand with its antiderivative, plus the conditions on the constants and on $x$ under which the formula holds. It extends the general list of integrals, which carries only the four headline radical rows, into full families with $x$, $x^2$, $\frac{1}{x}$ and higher powers attached.

The rows use short names for the radicals, so long formulas stay readable:

Every antiderivative carries an arbitrary constant $+ C$. The constant also absorbs fixed numbers such as $\ln a$, which is why two tables can print different-looking answers for the same integral.

How Do You Read The Conditions In Each Row?

The conditions tell you where a row is true, and skipping them is the most common way a correct formula gives a wrong answer. Read them in this order.

• Sign of the constant. In the $u$, $r$ and $s$ families, $a > 0$. If a problem has $\sqrt{9 - 4x^2}$, rewrite it as $2\sqrt{\frac{9}{4} - x^2}$ first, so the row applies with $a = \frac{3}{2}$.

• Domain of the root. A $u$ row only makes sense for $\lvert x\rvert \leq a$, and an $s$ row only for $\lvert x\rvert \geq a$. The domain and range of a function under the root fixes the interval for the answer too.

• Absolute values. A logarithm needs a positive argument. For $r$ rows, $x + r$ is always positive, so no absolute value is needed. For $s$ rows, $x + s$ is negative when $x \leq -a$, so the absolute value is required.

• Inverse function ranges. Rows with $\operatorname{arcsec}$ use $\operatorname{arcsec}\frac{\lvert x\rvert}{a}$, which keeps the formula valid on both sides, $x \geq a$ and $x \leq -a$.

Equivalent forms are not errors. For instance, $\ln\left(x + \sqrt{x^2+a^2}\right)$ and $\operatorname{arsinh}\frac{x}{a}$ differ by the constant $\ln a$, and both are correct antiderivatives.

What Are The Integrals Involving The Root Of A Linear Expression?

Integrals with a root of a linear expression all come from the substitution $t = \sqrt{ax+b}$, which turns them into polynomial or rational integrals in $t$. Here $L = \sqrt{ax+b}$ and $a \neq 0$.

The $\frac{1}{xL}$ row splits on the sign of $b$ because, after $t = L$, the denominator becomes $t^2 - b$. A positive $b$ factors into two linear terms and gives a logarithm; a negative $b$ leaves a sum of squares and gives an arctangent.

What Are The Integrals Involving The Root Of A Squared Minus X Squared?

These rows hold for $\lvert x\rvert < a$ (or $\lvert x\rvert \leq a$ where the integrand stays finite), with $u = \sqrt{a^2-x^2}$ and $a > 0$. Geometrically, $u$ is the upper half of a circle of radius $a$, so many of these integrals measure parts of a disc.

The rows with $\frac{1}{x}$ or $\frac{1}{x^2}$ also exclude $x = 0$. The first row is the area formula behind the disc: from $x = 0$ to $x = a$ it gives $\frac{\pi a^2}{4}$, a quarter of the circle's area, which makes a quick check on any copy of the table.

What Are The Integrals Involving The Root Of X Squared Plus A Squared?

These rows hold for every real $x$, with $r = \sqrt{x^2+a^2}$ and $a > 0$. A question that keeps appearing on exam-practice sites reads "The integral $\int\sqrt{x^2+a^2},dx$ is equal to", and the first row answers it directly.

Another frequently posted problem asks to evaluate $\int \frac{dx}{x^2\sqrt{1+x^2}}$. That is the $\frac{1}{x^2r}$ row with $a = 1$, so the answer is $-\frac{\sqrt{1+x^2}}{x} + C$ for $x \neq 0$. The logarithms in this family come from the hyperbolic functions route: with $x = a\sinh t$, the root becomes $a\cosh t$, and $\operatorname{arsinh}\frac{x}{a} = \ln\frac{x+r}{a}$.

What Are The Integrals Involving The Root Of X Squared Minus A Squared?

These rows hold for $\lvert x\rvert > a$ (or $\lvert x\rvert \geq a$ where the integrand stays finite), with $s = \sqrt{x^2-a^2}$ and $a > 0$. A Quora question asks "How to evaluate the integral $\int\sqrt{x^2-a^2},dx$"; the first row is the answer, and the second row is the piece it is built from.

Three Brainly questions land straight on these rows. "Evaluate the following integral: $\int \frac{dx}{\sqrt{x^2 - 484}}$, $x > 22$" is the $\frac{1}{s}$ row with $a = 22$, giving $\ln\left\lvert x + \sqrt{x^2-484}\right\rvert + C$. "Use the indicated substitution to evaluate the integral $\int \frac{1}{x\sqrt{x^2-361}},dx$" is the $\frac{1}{xs}$ row with $a = 19$, giving $\frac{1}{19}\operatorname{arcsec}\frac{\lvert x\rvert}{19} + C$. And $\int \frac{x^2}{\sqrt{x^2-7}},dx$, posted with "Determine an appropriate" substitution, is the $\frac{x^2}{s}$ row with $a = \sqrt{7}$: $\frac{x}{2}\sqrt{x^2-7} + \frac{7}{2}\ln\left\lvert x+\sqrt{x^2-7}\right\rvert + C$.

Compare the $\int r,dx$ and $\int s,dx$ rows. They differ only in the sign of the $\frac{a^2}{2}\ln$ term and in the absolute value, which is exactly where copied formulas go wrong. The $\operatorname{arcsec}$ rows need $\lvert x\rvert$ inside: without it, the derivative has the wrong sign for $x < -a$.

What Are The Integrals Involving The Root Of A General Quadratic?

A general quadratic under the root reduces to one of the three families above by completing the square, but the reduced results are worth listing directly. Here $R = \sqrt{ax^2+bx+c}$, $\Delta = 4ac - b^2$, and the quadratic is not a perfect square.

The $\int R,dx$ and $\int \frac{x,dx}{R}$ rows are reduction rows: they hand the work back to $\int \frac{dx}{R}$, so one of the first two rows always finishes the job. When $a < 0$, the root is real only between the two roots of the quadratic, which is why that row needs $b^2 - 4ac > 0$.

Which Other Irrational Integrals Are Worth Knowing?

A few results do not fit the five families but turn up often enough to keep beside them.

The last row is a reminder that rationalising the denominator is sometimes the whole method. Multiplying by $\sqrt{x+p} - \sqrt{x+q}$ turns the denominator into the constant $p - q$ (the same trade of a hard denominator for an easy one drives the Weierstrass substitution in trigonometric integrals).

How Are The Formulas In This List Derived?

Every row in this list comes from one of three techniques, and knowing which one produced a row lets you rebuild it when memory fails.

1. Direct substitution. Rows like $\int xu,dx = -\frac{u^3}{3} + C$ come from $w = a^2 - x^2$, because the derivative of the inside is already present. See integration by substitution for the pattern.

2. Trigonometric or hyperbolic substitution. Rows with arcsin, arcsec or $\ln(x + r)$ come from $x = a\sin\theta$, $a\sec\theta$ or $a\sinh t$. The trigonometric substitution page works through the reference triangles.

3. Integration by parts with a self-referencing step. Rows like $\int u,dx$ come from integration by parts, where the original integral reappears and is solved for.

Here is the third technique on the most-used row. Let $I = \int \sqrt{a^2-x^2},dx$ and integrate by parts with $dv = dx$:

$$I = x\sqrt{a^2-x^2} + \int \frac{x^2}{\sqrt{a^2-x^2}},dx$$

Write $x^2 = a^2 - \left(a^2 - x^2\right)$:

$$I = x\sqrt{a^2-x^2} + a^2\int \frac{dx}{\sqrt{a^2-x^2}} - \int \sqrt{a^2-x^2},dx$$

The last integral is $I$ again, so

$$2I = x\sqrt{a^2-x^2} + a^2\arcsin\frac{x}{a}$$

$$I = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\arcsin\frac{x}{a} + C$$

The same trick, with the sign of $a^2$ changed, produces the $\int r,dx$ and $\int s,dx$ rows.

Why Do Two Tables Give Different Answers For The Same Integral?

Two tables can disagree in form and still agree in fact, because antiderivatives are only fixed up to a constant. Before deciding a printed answer is wrong, differentiate both versions.

The common sources of difference are:

• Logarithm versus inverse hyperbolic. $\ln\left(x + \sqrt{x^2+a^2}\right)$ and $\operatorname{arsinh}\frac{x}{a}$ differ by $\ln a$.

• A constant inside the logarithm. $\ln\left\lvert\frac{x+s}{a}\right\rvert$ and $\ln\lvert x+s\rvert$ differ by $\ln a$.

• arcsin versus arccos. Since $\arcsin z + \arccos z = \frac{\pi}{2}$, the answers $\arcsin\frac{x}{a}$ and $-\arccos\frac{x}{a}$ differ by a constant.

• Rearranged logarithms. For $\int \frac{dx}{x\sqrt{x^2+x+1}}$, one valid answer is $\ln\left\lvert\frac{2\sqrt{x^2+x+1} - x - 2}{x}\right\rvert$ and another is $-\ln\left\lvert\frac{2 + x + 2\sqrt{x^2+x+1}}{x}\right\rvert$. They differ by $\ln 3$.

I should admit that the arcsec rows confused me longer than any other entry. Textbooks disagree on the range of arcsec for negative arguments, and the $\lvert x\rvert$ in these rows is the cleanest way to sidestep that disagreement.

Which Irrational Integrals Are Missing From Every List?

Some irrational integrals have no antiderivative made of elementary functions, so no list can contain them. The classic cases are square roots of cubic and quartic polynomials, such as $\int \sqrt{1+x^3},dx$ or $\int \frac{dx}{\sqrt{1-x^4}}$, which lead to elliptic integrals.

For these, a reference gives special-function forms or a numeric value instead. The definite integral $\int_{0}^{1}\sqrt{1+x^3},dx \approx 1.1114$ is computed numerically, and the NIST Digital Library of Mathematical Functions catalogues the elliptic integrals in its Chapter 19. If a search through every row turns up nothing, check the degree under the root before trying harder.

Examples Of Integrals Of Irrational Functions

Example 1

Find $\displaystyle\int \sqrt{16 - x^2},dx$, then use it to compute $\displaystyle\int_{0}^{4}\sqrt{16-x^2},dx$.

This example answers a Quora question, "How to integrate a square root with respect to x into a definite integral": find the antiderivative from the table, then evaluate it at the limits.

This is the $\int u,dx$ row with $a = 4$:

$$\int \sqrt{16-x^2},dx = \frac{x}{2}\sqrt{16-x^2} + 8\arcsin\frac{x}{4} + C$$

Evaluate from 0 to 4:

$$\left[\frac{x}{2}\sqrt{16-x^2} + 8\arcsin\frac{x}{4}\right]_{0}^{4} = (0 + 8\cdot\frac{\pi}{2}) - (0 + 0) = 4\pi$$

Check: a quarter of a disc with radius 4 has area $\frac{16\pi}{4} = 4\pi$.

Final answer: $\dfrac{x}{2}\sqrt{16-x^2} + 8\arcsin\dfrac{x}{4} + C$; the definite integral is $4\pi \approx 12.5664$

Example 2

Evaluate $\displaystyle\int_{-4}^{-3}\frac{dx}{\sqrt{x^2-4}}$.

The tempting route copies the $\frac{1}{s}$ row without the absolute value:

$$\left[\ln\left(x + \sqrt{x^2-4}\right)\right]_{-4}^{-3}$$

At $x = -3$, the argument is $-3 + \sqrt{5} \approx -0.7639$, a negative number. The logarithm of a negative number is undefined, yet the integrand $\frac{1}{\sqrt{x^2-4}}$ is positive and finite on the whole interval, so the integral clearly exists.

The fix is the condition printed with the row: for $\lvert x\rvert > a$, the antiderivative is $\ln\lvert x + s\rvert$. With $a = 2$:

$$\ln\lvert -3+\sqrt{5}\rvert - \ln\lvert -4+\sqrt{12}\rvert = \ln\left(3-\sqrt{5}\right) - \ln\left(4-2\sqrt{3}\right)$$

$$= \ln\frac{3-\sqrt{5}}{4-2\sqrt{3}} = \ln\frac{4+2\sqrt{3}}{3+\sqrt{5}}$$

The value is positive, as it must be for a positive integrand.

Final answer: $\ln\dfrac{4+2\sqrt{3}}{3+\sqrt{5}} \approx 0.3545$

Example 3

Find $\displaystyle\int \frac{x^2}{\sqrt{x^2+1}},dx$.

Use the $\frac{x^2}{r}$ row with $a = 1$:

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

No absolute value is needed, because $x + \sqrt{x^2+1} > 0$ for every real $x$.

Final answer: $\dfrac{x}{2}\sqrt{x^2+1} - \dfrac{1}{2}\ln\left(x+\sqrt{x^2+1}\right) + C$

Example 4

Find $\displaystyle\int \sqrt{x^2 + 6x + 13},dx$.

Complete the square:

$$x^2 + 6x + 13 = (x+3)^2 + 4$$

This is the $\int r,dx$ row with $a = 2$ and $x$ replaced by $x + 3$ (the shift does not change $dx$):

$$\int \sqrt{(x+3)^2+4},dx = \frac{x+3}{2}\sqrt{(x+3)^2+4} + \frac{4}{2}\ln\left(x+3+\sqrt{(x+3)^2+4}\right) + C$$

Final answer: $\dfrac{x+3}{2}\sqrt{x^2+6x+13} + 2\ln\left(x+3+\sqrt{x^2+6x+13}\right) + C$

Example 5

Find the arc length of the parabola $y = x^2$ from $x = 0$ to $x = 1$.

The arc length of a curve is $\int_{a}^{b}\sqrt{1 + \left(y'\right)^2},dx$, and $y' = 2x$:

$$L = \int_{0}^{1}\sqrt{1+4x^2},dx$$

Substitute $w = 2x$, $dx = \frac{dw}{2}$, with limits 0 to 2:

$$L = \frac{1}{2}\int_{0}^{2}\sqrt{w^2+1},dw$$

Apply the $\int r,dx$ row with $a = 1$:

$$L = \frac{1}{2}\left[\frac{w}{2}\sqrt{w^2+1} + \frac{1}{2}\ln\left(w+\sqrt{w^2+1}\right)\right]_{0}^{2}$$

$$= \frac{1}{2}\left[\sqrt{5} + \frac{1}{2}\ln\left(2+\sqrt{5}\right)\right] = \frac{\sqrt{5}}{2} + \frac{1}{4}\ln\left(2+\sqrt{5}\right)$$

Final answer: $\dfrac{\sqrt{5}}{2} + \dfrac{1}{4}\ln\left(2+\sqrt{5}\right) \approx 1.4789$

Example 6

Find $\displaystyle\int \frac{dx}{x\sqrt{4x+9}}$.

This is the $\frac{1}{xL}$ row with $a = 4$ and $b = 9 > 0$, so the logarithm version applies with $\sqrt{b} = 3$:

$$\int \frac{dx}{x\sqrt{4x+9}} = \frac{1}{3}\ln\left\lvert\frac{\sqrt{4x+9}-3}{\sqrt{4x+9}+3}\right\rvert + C$$

The result holds for $x > -\frac{9}{4}$ with $x \neq 0$.

Final answer: $\dfrac{1}{3}\ln\left\lvert\dfrac{\sqrt{4x+9}-3}{\sqrt{4x+9}+3}\right\rvert + C$

Why Do Integral Tables Exist At All?

"Nobody should have to derive the same integral twice." Integral tables exist because deriving a radical antiderivative takes several error-prone steps, while checking one takes a single differentiation. A trusted list moves the effort from deriving to checking.

• Reliability. A row that has been checked many times is safer than a fresh derivation done under time pressure.

• Speed in applied work. Engineers and physicists meet the same handful of radicals repeatedly, so a lookup saves real time.

• A record of conditions. A good table carries the domain and sign conditions that a hurried derivation tends to drop.

Leonhard Euler investigated in detail which integrals can be written with elementary functions in his Institutiones calculi integralis (1768 to 1770), one of the first systematic studies of the question this list answers row by row. In 1964 the US National Bureau of Standards published the Handbook of Mathematical Functions by Abramowitz and Stegun. Its modern successor, the NIST Digital Library of Mathematical Functions, was created as a complete revision of that handbook.

Where Are These Integrals Used In Real Life?

The square-root families appear whenever a length, a distance or a circle enters a calculation.

• Lengths of curves and cables. Arc length always carries $\sqrt{1 + (y')^2}$, and for parabolic and catenary shapes that is an $r$-family integral, as Example 5 shows.

• Areas and volumes of round shapes. The $u$ family measures areas under circles and ellipses, the starting point for areas of segments and volumes of spherical caps.

• Fields from straight sources. The electric field of a uniformly charged rod, or the magnetic field of a straight wire segment, leads to $\int \frac{dx}{\left(x^2+d^2\right)^{3/2}}$, the $\frac{1}{r^3}$ row.

• Motion under gravity. Time taken to fall between two heights under an inverse-square force produces radicals of the $\sqrt{\frac{a+x}{a-x}}$ type in the extra table.

Where Do Solutions Go Off The Rails With Irrational Integrals?

Mistake 1: Using A Row Outside Its Conditions

Where it slips in: Definite integrals over intervals where $x$ is negative, or where the root changes family.

Don't do this: Apply $\ln\left(x + \sqrt{x^2-a^2}\right)$ on an interval with $x < -a$, as in the tempting route of Example 2.

The correct way:Read the condition column before substituting numbers. The rusher goes straight from the integrand to the formula and skips that column, which is the only line that would have caught the negative logarithm.

Mistake 2: Misreading The Constant When X Squared Has A Coefficient

Where it slips in: Radicals like $\sqrt{9 - 4x^2}$ or $\sqrt{25x^2 + 16}$.

Don't do this: Read $a = 3$ from $\sqrt{9 - 4x^2}$ and use $\int u,dx$ directly.

The correct way: Factor the coefficient out first. Actually, factor it out of the whole radical, not just the $x^2$ term: $\sqrt{9-4x^2} = 2\sqrt{\frac{9}{4}-x^2}$, so $a = \frac{3}{2}$ and a factor of 2 sits outside. The first instinct on $\sqrt{9 - 4x^2}$ is to read $a = 3$ and stop; the 4 in front of $x^2$ is the part that gets lost. Students who always rewrite the radical in the exact table shape before matching never make this error.

Mistake 3: Copying The Wrong Sign Between The Plus And Minus Rows

Where it slips in: $\int \sqrt{x^2+a^2},dx$ and $\int \sqrt{x^2-a^2},dx$, which look almost identical.

Don't do this: Write $+\frac{a^2}{2}\ln\lvert x + s\rvert$ in the $\int s,dx$ answer.

The correct way: The $r$ row has $+\frac{a^2}{2}\ln$, and the $s$ row has $-\frac{a^2}{2}\ln$. Differentiate your final answer once to confirm the sign.

Mistake 4: Rejecting A Correct Answer Because It Looks Different

Where it slips in: Comparing your answer with an answer key that uses arsinh or a rearranged logarithm.

Don't do this: Rework a correct solution because the key prints $\operatorname{arsinh}\frac{x}{a}$ and you have $\ln\left(x + \sqrt{x^2+a^2}\right)$.

The correct way: Subtract the two answers. If the difference is a constant, both are right. The second-guesser loses the most time here, redoing work that was already correct.

Mistake 1 has a hardware cousin. In 1994, mathematician Thomas Nicely traced wrong division results on Intel's Pentium processor to its division lookup table: five of the 1,066 entries that the algorithm used held zero instead of their intended value. The Pentium FDIV bug led Intel to take a 475 million dollar charge. A table is only as good as each of its entries and the conditions under which you read them.

Practice Problems On The List Of Integrals Of Irrational Functions

1. Find $\displaystyle\int \frac{x^2}{\sqrt{4-x^2}},dx$.
   Answer to Question 1: $-\dfrac{x}{2}\sqrt{4-x^2} + 2\arcsin\dfrac{x}{2} + C$

2. Find $\displaystyle\int \frac{\sqrt{x^2-16}}{x},dx$ for $\lvert x\rvert > 4$.
   Answer to Question 2: $\sqrt{x^2-16} - 4\operatorname{arcsec}\dfrac{\lvert x\rvert}{4} + C$

3. Find $\displaystyle\int \frac{dx}{\left(x^2+9\right)^{3/2}}$.
   Answer to Question 3: $\dfrac{x}{9\sqrt{x^2+9}} + C$

4. Find $\displaystyle\int x\sqrt{2x+1},dx$.
   Answer to Question 4: $\dfrac{(3x-1)(2x+1)^{3/2}}{15} + C$

5. Evaluate $\displaystyle\int_{0}^{3}\sqrt{9-x^2},dx$.
   Answer to Question 5: $\dfrac{9\pi}{4} \approx 7.0686$

6. Find $\displaystyle\int \frac{dx}{\sqrt{x^2+4x+5}}$.
   Answer to Question 6: $\ln\left(x+2+\sqrt{x^2+4x+5}\right) + C$

The Short Version

• The list of integrals of irrational functions groups antiderivatives by radical: $\sqrt{ax+b}$, $\sqrt{a^2-x^2}$, $\sqrt{x^2+a^2}$, $\sqrt{x^2-a^2}$ and $\sqrt{ax^2+bx+c}$.

• Every row is valid only under its conditions, so read the sign of each constant and the interval for $x$ before using it.

• The $\sqrt{x^2-a^2}$ rows need an absolute value in the logarithm and $\lvert x\rvert$ inside arcsec.

• Answers that differ by a constant, such as a logarithm and arsinh, are equally correct.

• Square roots of cubics and quartics have no elementary antiderivative, so no list contains them.

Your Next Move

Work through practice problems 1 to 6, matching each one to a row before you compute; if the absolute value or the coefficient of $x^2$ trips you up, return to the conditions section and Example 2, then try again.

Book a free demo class to practise reading integral tables with a live trainer.

Read More

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• Definite integrals and how to evaluate them

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• Improper integrals and when they converge

• Table of integrals for every common function