Integration of Irrational Functions: Methods & Examples

What Is Integration Of Irrational Functions?

Integration of irrational functions is the set of methods for finding antiderivatives of algebraic functions that contain a root, such as $\sqrt{ax+b}$, $\sqrt[3]{x}$ or $\sqrt{ax^2+bx+c}$. A function is called irrational here when it cannot be written as a ratio of two polynomials, because a variable sits under a radical or carries a fractional power. The standard strategy, called rationalisation, picks a substitution that turns the integrand into a rational function of a new variable.

A question that keeps appearing on homework help sites reads: "Explain the integration of irrational algebraic functions." The short answer is the rationalisation idea above. In symbols, most school and first-year problems have the shape

$$\int R\left(x,\ \sqrt[n]{\frac{ax+b}{cx+d}}\right)dx \qquad \text{or} \qquad \int R\left(x,\ \sqrt{ax^2+bx+c}\right)dx,$$

where:

• $R$ is a rational function of its two arguments (built from them using only addition, subtraction, multiplication and division),

• $a, b, c, d$ are real constants, with $ad - bc \neq 0$ in the first form so the fraction is not a constant,

• $n \geq 2$ is a whole number, the index of the root.

The square root is by far the most common radical, so most of this page deals with it. Every answer below is an indefinite integral and carries a constant of integration $+ C$.

How Do You Choose The Right Method For An Irrational Integral?

Pick the method from the radical, not from the outside of the expression. A Quora question asks, "What is the process for simplifying an integral with a square root?" The process is to read what sits under the root, then match it to a row of the table below, which is the decision guide used throughout this page.

How Do You Integrate Roots Of Linear Expressions?

When the only radical is $\sqrt[n]{ax+b}$, make the whole radical the new variable. Set

$$t = \sqrt[n]{ax+b}, \qquad x = \frac{t^n - b}{a}, \qquad dx = \frac{n,t^{n-1}}{a},dt.$$

Every $x$ in the integrand becomes a polynomial in $t$, the root becomes $t$, and the integral turns into a rational integral. This is integration by substitution with one extra habit: write $dx$ in terms of $dt$ before touching anything else. Actually, the habit starts one line earlier: solve for $x$ first, because $dx$ comes straight from differentiating that expression.

Two variations come up often.

• Several roots of the same expression. For $\sqrt{x}$ and $\sqrt[3]{x}$ together, use $t = x^{1/6}$, since 6 is the least common multiple of 2 and 3. Then $\sqrt{x} = t^3$ and $\sqrt[3]{x} = t^2$, both whole powers.

• A root of a fraction. For $\sqrt[n]{\frac{ax+b}{cx+d}}$, set $t^n = \frac{ax+b}{cx+d}$ and solve for $x$. The result is a rational function of $t$, and so is $dx$.

If the new integrand is a rational function with a higher-degree numerator, divide first, then use integration by partial fractions on what remains. Example 3 below runs this full chain.

When Is Plain Substitution Enough?

Plain substitution is enough whenever the derivative of the expression under the root already sits outside it. For instance, $\int \frac{x}{\sqrt{x^2+4}},dx$ has $2x$ (up to a factor) next to $x^2+4$, so $u = x^2 + 4$ gives $\frac{1}{2}\int u^{-1/2},du = \sqrt{x^2+4} + C$.

Check for this pattern before reaching for anything heavier. A Brainly question asks for trigonometric substitution on $\int \frac{x^3}{\sqrt{x^2+4}},dx$, yet $u = x^2 + 4$ (with $x^2 = u - 4$) is shorter and gives $\frac{1}{3}\left(x^2+4\right)^{3/2} - 4\sqrt{x^2+4} + C$.

The pattern also explains a Quora question: "Why can't we solve the integral of the square root of x2+1 using the fundamental theorem of calculus? Why can't I use the substitution method?" In $\int \sqrt{x^2+1},dx$ there is no $x$ outside the root to cancel the $2x$ from $du$, so $u = x^2 + 1$ stalls. The fundamental theorem still applies once an antiderivative is known; finding that antiderivative needs $x = \tan\theta$ or $x = \sinh t$, which gives the $\sqrt{x^2+a^2}$ row of the standard table.

How Does Trigonometric Substitution Handle The Three Square Root Forms?

The three pure quadratic radicals each match a Pythagorean identity, and the matching substitution removes the root completely. The full method, reference triangles and back-substitution are worked through on the trigonometric substitution page, so here is only the map:

There is a second route that many textbooks skip. Hyperbolic functions satisfy $\cosh^2 u - \sinh^2 u = 1$, so $x = a\sinh u$ handles $\sqrt{a^2+x^2}$ and $x = a\cosh u$ handles $\sqrt{x^2-a^2}$. That route explains why the answers for those two radicals contain a natural logarithm rather than an inverse trig function: $\operatorname{arsinh}\frac{x}{a} = \ln\left(x + \sqrt{x^2+a^2}\right) - \ln a$.

What Are The Standard Forms Of Irrational Integrals?

Six results cover most problems once the quadratic has been reduced to one of the three pure forms. Here $a > 0$, and each result holds on the interval where its root is real.

Each row can be checked by differentiating. For the first row, the derivatives of inverse trigonometric functions give $\frac{d}{dx}\arcsin\frac{x}{a} = \frac{1}{\sqrt{a^2-x^2}}$ directly. The absolute value in the $\sqrt{x^2-a^2}$ rows matters: for $x < -a$, the quantity $x + \sqrt{x^2-a^2}$ is negative, and the logarithm needs its absolute value.

How Do You Integrate A Linear Expression Over The Square Root Of A Quadratic?

Write the linear numerator as a multiple of the quadratic's derivative plus a constant, then split. Question pages titled "Integral of polynomial/sqrt(quadratic) or $\int \frac{px+q}{\sqrt{ax^2+bx+c}},dx$" are among the most searched shapes in this topic.

The method has three steps.

1. Match the derivative. Find $\lambda$ and $\mu$ with $px + q = \lambda(2ax + b) + \mu$. Comparing coefficients gives $\lambda = \dfrac{p}{2a}$ and $\mu = q - \dfrac{pb}{2a}$.

2. Split the integral. The first piece has the form $\dfrac{f'(x)}{\sqrt{f(x)}}$, which integrates to $2\sqrt{f(x)}$. The second piece is $\mu\displaystyle\int \frac{dx}{\sqrt{ax^2+bx+c}}$.

3. Complete the square in the second piece and read off a standard form.

Put together,

$$\int \frac{px+q}{\sqrt{ax^2+bx+c}},dx = \frac{p}{a}\sqrt{ax^2+bx+c} + \mu\int \frac{dx}{\sqrt{ax^2+bx+c}}.$$

The last integral needs completing the square: $ax^2 + bx + c$ becomes $a\left[(x+h)^2 \pm k^2\right]$, and one of the standard forms in the table above finishes the job.

What If The Square Root Is In The Numerator?

A Quora thread asks, "How do I find the integral of type (px+q)(ax²+bx+c)^(1/2) dx?" The same split works. With $px + q = \lambda(2ax+b) + \mu$,

$$\int (px+q)\sqrt{ax^2+bx+c},dx = \frac{2\lambda}{3}\left(ax^2+bx+c\right)^{3/2} + \mu\int \sqrt{ax^2+bx+c},dx,$$

and the last integral, after completing the square, is one of the three $\sqrt{\ \ }$ rows of the standard table.

What About One Over A Linear Factor Times A Square Root?

A homework question that turns up often asks for $\int \frac{1}{(px+q)\sqrt{ax^2+bx+c}},dx$. The move here is the reciprocal substitution $t = \frac{1}{px+q}$. It turns the linear factor into a power of $t$ and leaves a single square root of a quadratic in $t$.

For example, with $x > -1$ and $t = \frac{1}{x+1}$, the integral $\int \frac{dx}{(x+1)\sqrt{x^2+2x+2}}$ becomes $-\int \frac{dt}{\sqrt{1+t^2}}$, so the answer is $-\operatorname{arsinh}\frac{1}{x+1} + C$. The restriction $x > -1$ matters: it keeps $t > 0$, which the simplification $\sqrt{t^2} = t$ needs.

What Are Euler Substitutions And When Do You Use Them?

Euler substitutions turn any integral of the form $\int R\left(x, \sqrt{ax^2+bx+c}\right)dx$ into the integral of a rational function. They are the general tool. Trigonometric substitution is faster for pure forms, but Euler substitutions work even when nothing else fits.

The three substitutions, with $y = \sqrt{ax^2+bx+c}$:

• First substitution, when $a > 0$: set $\sqrt{ax^2+bx+c} = t - \sqrt{a},x$ (or $t + \sqrt{a},x$). Squaring cancels the $ax^2$ terms, so $x$ comes out as a rational function of $t$.

• Second substitution, when $c > 0$: set $\sqrt{ax^2+bx+c} = xt + \sqrt{c}$ (or $xt - \sqrt{c}$). Squaring cancels the constant, and dividing by $x$ leaves $x$ rational in $t$.

• Third substitution, when the quadratic has real roots $\alpha$ and $\beta$: set $\sqrt{a(x-\alpha)(x-\beta)} = t(x-\alpha)$. Squaring and dividing by $x - \alpha$ leaves a linear equation in $x$.

A Physics Forums thread titled "Integration using Euler Substitution" asks for "a derivation or justification of Euler's substitution formulas", and in particular "when to use t+x as opposed to t−x." The justification is geometric. The equation $y^2 = ax^2 + bx + c$ is a conic. Any line through a known point of that conic meets it in exactly one more point, and the coordinates of that second point are rational functions of the line's slope. Each Euler substitution is one choice of known point:

• The first substitution uses a "point at infinity" in the direction of the asymptote $y = \pm\sqrt{a},x$, which exists only when $a > 0$.

• The second uses the point $(0, \sqrt{c})$, which lies on the curve only when $c > 0$.

• The third uses the point $(\alpha, 0)$ where the curve crosses the $x$-axis.

So the plus or minus sign is a free choice. Both signs give valid rational substitutions. Pick the one that keeps the algebra shorter, which is usually the sign that makes the denominator of $x(t)$ simpler. The sign rule took me a while to accept for exactly this reason: textbooks print one sign as if it were the rule, when it is only a convenience.

Which Irrational Integrals Have No Elementary Answer?

Not every irrational function has an antiderivative made of elementary functions, and no substitution will change that. The cleanest test covers the binomial differential $x^m(a + bx^n)^p,dx$ with rational $m, n, p$. Its integral is elementary only when at least one of these is an integer:

$$p, \qquad \frac{m+1}{n}, \qquad \frac{m+1}{n} + p.$$

Take $\int \sqrt{1 + x^3},dx$. Here $m = 0$, $n = 3$, $p = \frac{1}{2}$, so the three numbers are $\frac{1}{2}$, $\frac{1}{3}$ and $\frac{5}{6}$. None is an integer, so this integral is non-elementary (it is an elliptic integral).

A definite value still exists: $\int_{0}^{1} \sqrt{1+x^3},dx \approx 1.1114$, found numerically. Compare $\int x^2\sqrt{1+x^3},dx$, where $\frac{m+1}{n} = 1$ is an integer, and the substitution $u = 1 + x^3$ finishes it in one line.

Square roots of cubics and quartics are the usual source of these dead ends (computer algebra systems settle the general question with the Risch algorithm). Recognising them saves you from hunting for a substitution that does not exist.

Examples Of Integration Of Irrational Functions

Example 1

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

Let $t = \sqrt{x+1}$, so $x = t^2 - 1$ and $dx = 2t,dt$:

$$\int x\sqrt{x+1},dx = \int (t^2-1)\cdot t\cdot 2t,dt$$

$$= \int \left(2t^4 - 2t^2\right)dt$$

$$= \frac{2t^5}{5} - \frac{2t^3}{3} + C$$

Substitute back $t = \sqrt{x+1}$.

Final answer: $\dfrac{2}{5}(x+1)^{5/2} - \dfrac{2}{3}(x+1)^{3/2} + C$

Example 2

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

The tempting move is to match the shape $\frac{1}{\sqrt{\text{something}}}$ to the arcsin formula and write

$$\int \frac{dx}{\sqrt{x^2+9}} = \arcsin\frac{x}{3} + C \qquad \text{(tempting, but wrong)}$$

Differentiate that answer to check it:

$$\frac{d}{dx}\arcsin\frac{x}{3} = \frac{1}{\sqrt{9-x^2}}$$

That is the derivative for $\sqrt{9 - x^2}$, not $\sqrt{x^2 + 9}$. The guess also breaks for $\lvert x\rvert > 3$, where arcsin is undefined, even though $\sqrt{x^2+9}$ is defined for every $x$.

The sign inside the root decides the form. A sum of squares calls for $x = 3\tan\theta$ or $x = 3\sinh u$, and the standard table gives

$$\int \frac{dx}{\sqrt{x^2+9}} = \ln\left(x + \sqrt{x^2+9}\right) + C$$

Check: the derivative of $\ln\left(x+\sqrt{x^2+9}\right)$ is $\dfrac{1 + \frac{x}{\sqrt{x^2+9}}}{x+\sqrt{x^2+9}} = \dfrac{1}{\sqrt{x^2+9}}$.

Final answer: $\ln\left(x + \sqrt{x^2+9}\right) + C$, equivalently $\operatorname{arsinh}\dfrac{x}{3} + C$

Example 3

Find $\displaystyle\int \frac{dx}{\sqrt{x} + \sqrt[3]{x}}$ for $x > 0$.

The indices are 2 and 3, with LCM 6. Let $t = x^{1/6}$, so $x = t^6$, $dx = 6t^5,dt$, $\sqrt{x} = t^3$ and $\sqrt[3]{x} = t^2$:

$$\int \frac{6t^5,dt}{t^3 + t^2} = 6\int \frac{t^3}{t+1},dt$$

Divide the polynomials:

$$\frac{t^3}{t+1} = t^2 - t + 1 - \frac{1}{t+1}$$

Integrate term by term:

$$6\left(\frac{t^3}{3} - \frac{t^2}{2} + t - \ln(t+1)\right) + C = 2t^3 - 3t^2 + 6t - 6\ln(t+1) + C$$

Final answer: $2\sqrt{x} - 3\sqrt[3]{x} + 6\sqrt[6]{x} - 6\ln\left(1 + \sqrt[6]{x}\right) + C$

Example 4

Find $\displaystyle\int \frac{2x+3}{\sqrt{x^2+2x+5}},dx$.

The derivative of $x^2 + 2x + 5$ is $2x + 2$, so write $2x + 3 = 1\cdot(2x+2) + 1$:

$$\int \frac{2x+2}{\sqrt{x^2+2x+5}},dx + \int \frac{dx}{\sqrt{x^2+2x+5}}$$

The first integral is $2\sqrt{x^2+2x+5}$. For the second, complete the square:

$$x^2 + 2x + 5 = (x+1)^2 + 4$$

$$\int \frac{dx}{\sqrt{(x+1)^2+4}} = \ln\left(x + 1 + \sqrt{x^2+2x+5}\right) + C$$

Final answer: $2\sqrt{x^2+2x+5} + \ln\left(x + 1 + \sqrt{x^2+2x+5}\right) + C$

Example 5

Find $\displaystyle\int \sqrt{3 - 2x - x^2},dx$.

Complete the square with the negative leading coefficient handled first:

$$3 - 2x - x^2 = 3 - (x^2 + 2x) = 4 - (x+1)^2$$

This is the $\sqrt{a^2 - u^2}$ form with $a = 2$ and $u = x + 1$:

$$\int \sqrt{4 - (x+1)^2},dx = \frac{x+1}{2}\sqrt{4-(x+1)^2} + \frac{4}{2}\arcsin\frac{x+1}{2} + C$$

The result holds for $-3 \leq x \leq 1$, where the root is real.

Final answer: $\dfrac{x+1}{2}\sqrt{3-2x-x^2} + 2\arcsin\dfrac{x+1}{2} + C$

Example 6

Use Euler's second substitution to find $\displaystyle\int \frac{dx}{x\sqrt{x^2+x+1}}$.

Here $c = 1 > 0$. Set $\sqrt{x^2+x+1} = tx + 1$ and square:

$$x^2 + x + 1 = t^2x^2 + 2tx + 1$$

Cancel the 1, divide by $x$:

$$x + 1 = t^2x + 2t \quad\Longrightarrow\quad x = \frac{2t-1}{1-t^2}$$

Then

$$\sqrt{x^2+x+1} = tx + 1 = \frac{t^2-t+1}{1-t^2}, \qquad dx = \frac{2\left(t^2-t+1\right)}{\left(1-t^2\right)^2},dt$$

Substitute and simplify:

$$\int \frac{dx}{x\sqrt{x^2+x+1}} = \int \frac{2,dt}{2t-1} = \ln\lvert 2t-1\rvert + C$$

Back-substitute $t = \dfrac{\sqrt{x^2+x+1} - 1}{x}$:

$$2t - 1 = \frac{2\sqrt{x^2+x+1} - x - 2}{x}$$

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

Some tables print $-\ln\left\lvert \frac{2 + x + 2\sqrt{x^2+x+1}}{x}\right\rvert + C$ instead. The two answers differ by the constant $\ln 3$, so both are correct. A Brainly question makes the same point about integrals in general: "The indefinite integral can be found in more than one way." Different methods may print different forms, and a constant difference is the test that they agree.

Why Do We Rationalise Instead Of Inventing New Rules?

"Trade the root for a fraction, then use what already works." That idea is the reason this topic exists, and it is economical. Integration has a complete method for rational functions: divide, then use partial fractions, and the answer is always built from polynomials, logarithms and arctangents. Rather than inventing a separate rulebook for every radical, mathematicians looked for substitutions that move a radical problem into that solved territory.

• It reuses a finished theory. Once the integrand is rational, nothing new is needed. The hard thinking happens in choosing the substitution.

• It tells you when to stop. If no substitution can rationalise the integrand, the binomial test above often proves that no elementary answer exists, which is useful knowledge in itself.

• It feeds measurement. Arc length, surface area and many physics integrals produce square roots by their nature, so a reliable way to remove them was a practical need.

Leonhard Euler made this a systematic programme in his Institutiones calculi integralis (1768 to 1770), a two-volume work that investigated which integrals can be expressed in elementary functions. The substitutions that carry his name come from that tradition.

Where Is Integration Of Irrational Functions Used In Real Life?

Square roots enter real formulas whenever a distance or a length is involved, so these integrals turn up across applied work.

• Arc length. The length of a curve $y = f(x)$ is $\int_{a}^{b}\sqrt{1 + \left(f'(x)\right)^2},dx$. For the parabola $y = x^2$ from 0 to 1, that is $\int_{0}^{1}\sqrt{1+4x^2},dx \approx 1.4789$, a direct use of the $\sqrt{x^2+a^2}$ form. The arc length of a curve page builds this formula.

• Hanging cables and chains. The length of a cable that hangs in a catenary comes from a root of the form $\sqrt{1+\sinh^2}$, which simplifies thanks to the hyperbolic identity.

• Electric and gravitational fields. The field of a straight line charge or rod produces integrals like $\int \frac{dx}{\left(x^2+d^2\right)^{3/2}}$, a power of the $\sqrt{x^2+a^2}$ radical.

• Pendulums and orbits. Large-swing pendulum periods and orbit timing lead to square roots of cubics and quartics, which are the elliptic integrals catalogued by NIST in the Digital Library of Mathematical Functions. These are exactly the cases the binomial test flags as non-elementary.

Where Do Students Trip Up On Integration Of Irrational Functions?

Mistake 1: Forgetting To Convert The Differential

Where it slips in: The first line of a linear-root substitution such as $t = \sqrt{x+1}$.

Don't do this: Replace the root with $t$ and the $x$ terms with $t^2 - 1$, but leave $dx$ as $dt$. The factor $2t$ disappears and every later step is wrong.

The correct way: Write the three-line block $t = \dots$, $x = \dots$, $dx = \dots,dt$ before substituting anything. The rusher skips this block, and that is precisely where the missing factor comes from.

Mistake 2: Choosing The Form From The Outline, Not The Sign

Where it slips in: Any $\frac{1}{\sqrt{\ \ }}$ integral, such as $\int \frac{dx}{\sqrt{x^2+9}}$.

Don't do this: Reach for arcsin because the expression "looks like" $\frac{1}{\sqrt{a^2-x^2}}$.

The correct way: Read the sign between the squares first. The first instinct on $\frac{1}{\sqrt{x^2+9}}$ is arcsin, and the only thing that should decide is whether $x^2$ is subtracted or added. Minus $x^2$ gives arcsin; plus $x^2$ gives a logarithm. The memoriser who learned the table as pictures rather than signs falls into this one most often.

Mistake 3: Completing The Square Wrongly With A Negative Leading Term

Where it slips in: Quadratics like $3 - 2x - x^2$ or $5 - 4x - x^2$.

Don't do this: Write $3 - 2x - x^2 = (x-1)^2 + 2$ or any form where the sign of $x^2$ has flipped.

The correct way: Factor out the minus sign first: $3 - (x^2 + 2x) = 3 - \left[(x+1)^2 - 1\right] = 4 - (x+1)^2$. Expand your answer once to check that it returns the original quadratic. Students who always expand back catch this error before it costs anything.

Mistake 4: Dropping The Absolute Value Or The Domain

Where it slips in: Answers with $\ln\lvert x + \sqrt{x^2-a^2}\rvert$, or substitutions that use $\sqrt{t^2} = t$.

Don't do this: Write $\ln\left(x + \sqrt{x^2-a^2}\right)$ and then evaluate it at $x = -3$ with $a = 2$, where the argument is negative.

The correct way: Keep the absolute value, and state the interval where each step is valid, as Example 5 and the reciprocal substitution above do.

Mistake 2 has a famous real-world cousin: judging a curve by its outline. In Two New Sciences (1638), Galileo wrote that a hanging chain is only approximately a parabola. Joachim Jungius proved it is not a parabola at all (published in 1669), and the exact curve, the catenary, was derived in 1691 by Leibniz, Huygens and Johann Bernoulli. Two curves that look alike, like two radicals that look alike, can need completely different mathematics.

Practice Problems On Integration Of Irrational Functions

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

2. Find $\displaystyle\int \frac{\sqrt{x}}{1+x},dx$ for $x > 0$.
   Answer to Question 2: $2\sqrt{x} - 2\arctan\sqrt{x} + C$

3. Find $\displaystyle\int \frac{x+1}{\sqrt{x^2+4x+8}},dx$.
   Answer to Question 3: $\sqrt{x^2+4x+8} - \ln\left(x + 2 + \sqrt{x^2+4x+8}\right) + C$

4. Find $\displaystyle\int \frac{dx}{\sqrt{x^2-6x+5}}$, where $\lvert x-3\rvert > 2$.
   Answer to Question 4: $\ln\left\lvert x - 3 + \sqrt{x^2-6x+5}\right\rvert + C$

5. Find $\displaystyle\int \frac{dx}{\sqrt{5-4x-x^2}}$.
   Answer to Question 5: $\arcsin\dfrac{x+2}{3} + C$

6. Find $\displaystyle\int \frac{x^3}{\sqrt{1+x^2}},dx$.
   Answer to Question 6: $\dfrac{1}{3}\left(1+x^2\right)^{3/2} - \sqrt{1+x^2} + C$

Key Takeaways

• Integration of irrational functions works by choosing a substitution that turns the radical into a rational function, then integrating that.

• A root of a linear expression becomes the new variable, and several roots of $x$ use the least common multiple of their indices.

• Linear over a square root of a quadratic splits into a derivative piece and a completed-square standard form.

• Euler substitutions rationalise any $R(x, \sqrt{ax^2+bx+c})$, and each one comes from a line through a known point of a conic.

• The sign inside the root, not its outline, decides between arcsin and a logarithm.

Where To Go From Here

Work through practice problems 1 to 6 in order; if the $dx$ conversion or the sign of the square trips you up, return to the section on roots of linear expressions or the standard forms table and rework Example 1 or Example 2 before trying again.

Book a free demo class to practise these substitutions with a live trainer.

Read More

• Methods of integration compared side by side

• Integration of rational functions with partial fractions

• List of integrals for standard functions

• Table of integrals for quick lookup

• Integration by parts with worked examples

• Applications of integration in geometry and physics