Tutorials › Calculus II › Rationalizing √(x² + a²)

Home  /  Tutorials  /  Rationalizing Substitutions

Rationalizing Substitutions · Tutorial 236 of 998

Rationalizing √(x² + a²)

Learn how this substitution turns the radical and differential into rational expressions, how to choose its branch, and how to check the resulting antiderivatives.

Intermediate 9 min read

What You'll Learn

  • Derive the radical and differential after setting x = 2at/(1−t²)
  • Identify the branch −1 < t < 1 and its range in x
  • Recover the substitution parameter from x and the radical
  • Apply the substitution to three representative integrals
  • Check transformed antiderivatives and avoid common sign and factor errors

Why Use This Substitution?

When an integral contains \(\sqrt{x^2+a^2}\), the radical may obstruct a direct algebraic calculation. A useful change of variable replaces both \(x\) and the radical by rational functions of a new parameter. The differential \(dx\) becomes rational as well, so an integrand built rationally from \(x\) and \(\sqrt{x^2+a^2}\) can often be converted into a rational function of that parameter.

Throughout, assume \(a>0\). The substitution we will use is

$$ x=\frac{2at}{1-t^2}. $$

The branch \(-1<t<1\) is essential: it gives one parameter value for every real \(x\), and it ensures that the denominator \(1-t^2\) is positive. We will derive the transformed radical and differential, explain how the branch works, and use the substitution in several integrals.

Deriving the Radical and Differential

Start with \(x=2at/(1-t^2)\), where \(a>0\) and \(-1<t<1\). Squaring the expression for \(x\) and adding \(a^2\) gives

$$ \begin{aligned} x^2+a^2 &=\frac{4a^2t^2}{(1-t^2)^2}+a^2\\ &=\frac{4a^2t^2+a^2(1-2t^2+t^4)}{(1-t^2)^2}\\ &=\frac{a^2(1+2t^2+t^4)}{(1-t^2)^2}\\ &=\frac{a^2(1+t^2)^2}{(1-t^2)^2}. \end{aligned} $$

Because \(a>0\), \(1+t^2>0\), and \(1-t^2>0\) on this branch, taking the positive square root yields

$$ \sqrt{x^2+a^2}=\frac{a(1+t^2)}{1-t^2}. $$

Differentiate the expression for \(x\) with respect to \(t\). The quotient rule gives

$$ \begin{aligned} \frac{dx}{dt} &=\frac{2a(1-t^2)-2at(-2t)}{(1-t^2)^2}\\ &=\frac{2a(1-t^2+2t^2)}{(1-t^2)^2}\\ &=\frac{2a(1+t^2)}{(1-t^2)^2}. \end{aligned} $$

Thus the differential is

$$ dx=\frac{2a(1+t^2)}{(1-t^2)^2}\,dt. $$
Key substitution formulas: For \(a>0\), \(-1<t<1\), and \(x=2at/(1-t^2)\), \[ \sqrt{x^2+a^2}=\frac{a(1+t^2)}{1-t^2}, \qquad dx=\frac{2a(1+t^2)}{(1-t^2)^2}\,dt. \]

Why This Branch Covers the Real Line

To use a substitution in an integral, it matters whether the chosen parameter range actually corresponds to the \(x\)-values of interest. Define \[ x(t)=\frac{2at}{1-t^2} \] for \(-1<t<1\). Its derivative is \(2a(1+t^2)/(1-t^2)^2\), which is positive because every factor in that expression is positive. Therefore \(x(t)\) is strictly increasing on the interval.

At the endpoints, the values of \(x(t)\) grow without bound in opposite directions:

$$ \lim_{t\to-1^+}\frac{2at}{1-t^2}=-\infty, \qquad \lim_{t\to1^-}\frac{2at}{1-t^2}=+\infty. $$

The function is continuous and strictly increasing, with these endpoint limits. It therefore takes each real value of \(x\) exactly once. We can also solve explicitly for the parameter. Let \(s=\sqrt{x^2+a^2}\). Then

$$ t=\frac{x}{s+a}=\frac{x}{\sqrt{x^2+a^2}+a}. $$

This expression is always in \((-1,1)\): since \(s>|x|\) for \(a>0\), we have \(|x|<s+a\). It also gives \(t=0\) when \(x=0\). Substituting this \(t\) into \(2at/(1-t^2)\) recovers \(x\), so it is the inverse parameter on the chosen branch.

Branch condition: For \(a>0\), the substitution \(x=2at/(1-t^2)\) with \(-1<t<1\) is increasing and covers all real \(x\). On this branch, \(1-t^2\) is positive, so the displayed formula for the square root has the correct positive sign.

Worked Example: Integrating the Reciprocal Radical

Worked Example: \(\int dx/\sqrt{x^2+a^2}\)

Assume \(a>0\). Apply the substitution formulas to the integrand:

$$ \begin{aligned} \int\frac{dx}{\sqrt{x^2+a^2}} &=\int \frac{\dfrac{2a(1+t^2)}{(1-t^2)^2}} {\dfrac{a(1+t^2)}{1-t^2}}\,dt\\ &=\int\frac{2}{1-t^2}\,dt. \end{aligned} $$

Use the partial-fraction identity \(2/(1-t^2)=1/(1-t)+1/(1+t)\). Since \(-1<t<1\), both \(1-t\) and \(1+t\) are positive, so

$$ \begin{aligned} \int\frac{2}{1-t^2}\,dt &=\int\left(\frac{1}{1-t}+\frac{1}{1+t}\right)\,dt\\ &=-\ln(1-t)+\ln(1+t)+C\\ &=\ln\left(\frac{1+t}{1-t}\right)+C. \end{aligned} $$

To express the answer in \(x\), put \(s=\sqrt{x^2+a^2}\) and \(t=x/(s+a)\). Then

$$ \begin{aligned} \frac{1+t}{1-t} &=\frac{s+a+x}{s+a-x}\\ &=\frac{(s+a+x)^2}{(s+a-x)(s+a+x)}\\ &=\frac{(s+a+x)^2}{2a(s+a)}\\ &=\frac{s+x}{a}. \end{aligned} $$

The last equality follows from \((s+a+x)^2=2(s+a)(s+x)\), using \(s^2=x^2+a^2\). Therefore an antiderivative is \(\ln((x+\sqrt{x^2+a^2})/a)+C\). Check by differentiation: the derivative of \(\ln(s+x)\) is \[ \frac{1}{s+x}\left(\frac{x}{s}+1\right)=\frac{1}{s}, \] and the constant \(-\ln a\) has derivative zero. This gives the original integrand.

Worked Example: Integrating \(1/(x^2+a^2)\)

Worked Example: \(\int dx/(x^2+a^2)\)

Again take \(a>0\). First express the denominator using \(t\):

$$ x^2+a^2=\frac{a^2(1+t^2)^2}{(1-t^2)^2}. $$

Now include \(dx\) and simplify every factor:

$$ \begin{aligned} \int\frac{dx}{x^2+a^2} &=\int \frac{(1-t^2)^2}{a^2(1+t^2)^2} \frac{2a(1+t^2)}{(1-t^2)^2}\,dt\\ &=\frac{2}{a}\int\frac{dt}{1+t^2}\\ &=\frac{2}{a}\arctan(t)+C. \end{aligned} $$

Substitute \(t=x/(\sqrt{x^2+a^2}+a)\). To check the answer directly, write \(s=\sqrt{x^2+a^2}\). Differentiating the inverse parameter gives

$$ \begin{aligned} \frac{dt}{dx} &=\frac{(s+a)-x(x/s)}{(s+a)^2}\\ &=\frac{a+a^2/s}{(s+a)^2}\\ &=\frac{a}{s(s+a)}. \end{aligned} $$

Also, \(1+t^2=2s/(s+a)\). Hence

$$ \frac{d}{dx}\left(\frac{2}{a}\arctan t\right) =\frac{2}{a}\frac{1}{1+t^2}\frac{dt}{dx} =\frac{2}{a}\frac{s+a}{2s}\frac{a}{s(s+a)} =\frac{1}{s^2} =\frac{1}{x^2+a^2}. $$

The derivative agrees with the original integrand.

Worked Example: Integrating \(x/\sqrt{x^2+a^2}\)

Worked Example: \(\int x\,dx/\sqrt{x^2+a^2}\)

For \(a>0\), the transformed factor \(x/\sqrt{x^2+a^2}\) is

$$ \frac{x}{\sqrt{x^2+a^2}} =\frac{\dfrac{2at}{1-t^2}}{\dfrac{a(1+t^2)}{1-t^2}} =\frac{2t}{1+t^2}. $$

Multiplying by the transformed differential gives

$$ \begin{aligned} \int\frac{x}{\sqrt{x^2+a^2}}\,dx &=\int\frac{2t}{1+t^2} \frac{2a(1+t^2)}{(1-t^2)^2}\,dt\\ &=\int\frac{4at}{(1-t^2)^2}\,dt\\ &=\frac{2a}{1-t^2}+C. \end{aligned} $$

For the last integration, the derivative of \(1/(1-t^2)\) is \(2t/(1-t^2)^2\), so the derivative of \(2a/(1-t^2)\) is \(4at/(1-t^2)^2\). In terms of \(x\),

$$ \frac{2a}{1-t^2} =\frac{2a(s+a)^2}{(s+a)^2-x^2} =\frac{2a(s+a)^2}{2a(s+a)} =s+a, $$

where \(s=\sqrt{x^2+a^2}\). Thus the antiderivative is \(\sqrt{x^2+a^2}+C\). Its derivative is \(x/\sqrt{x^2+a^2}\), as required.

Why the Substitution Rationalizes

The construction succeeds because both \(x\) and the radical have rational expressions in \(t\), and their quotient and products with \(dx\) are rational as well. For example, the defining identities give

$$ x=\frac{2at}{1-t^2}, \qquad \sqrt{x^2+a^2}=\frac{a(1+t^2)}{1-t^2}, \qquad dx=\frac{2a(1+t^2)}{(1-t^2)^2}\,dt. $$

Consequently, if an integrand is a rational expression in \(x\) and \(\sqrt{x^2+a^2}\), substitution replaces each of those quantities by rational functions of \(t\). After multiplying by \(dx\), the resulting integrand is rational in \(t\), wherever the original integrand and the substitution are defined. This is the practical reason the parameter is useful: it turns an algebraic radical problem into a rational integration problem.

The branch also explains why no absolute-value correction appears in the radical formula. In general, taking a square root of a squared denominator introduces its absolute value. Here \(|1-t^2|=1-t^2\), precisely because \(-1<t<1\). Choosing another parameter interval without reconsidering that sign can produce an incorrect expression for the positive square root.

A Common Mistake: Losing a Factor in \(dx\)

A frequent error is to differentiate \(2at/(1-t^2)\) incompletely and omit the factor \(1+t^2\). For example, the incorrect calculation

$$ dx\ \text{incorrectly written as}\ \frac{2a}{(1-t^2)^2}\,dt $$

would turn \(\int dx/\sqrt{x^2+a^2}\) into \(\int 2/(1+t^2)\,dt\). That answer does not follow from the substitution. The quotient rule has two contributions in the numerator:

$$ 2a(1-t^2)-2at(-2t)=2a(1+t^2). $$

The correct differential therefore includes \(1+t^2\), and the correct transformed integral is \(\int 2/(1-t^2)\,dt\). The resulting antiderivative differentiates to \(1/\sqrt{x^2+a^2}\), while the expression from the incorrect differential does not. Keeping the full derivative visible before simplifying helps prevent this error.

Using the Substitution Reliably

When applying this method, record the assumptions and the branch before simplifying. The formulas above use \(a>0\) and \(-1<t<1\), and the inverse relation is \(t=x/(\sqrt{x^2+a^2}+a)\). If an integral is definite, its endpoints in \(x\) can be converted with this inverse relation; the increasing branch preserves their order. For an indefinite integral, the final expression can be written using \(t\) or the inverse formula in \(x\), with an arbitrary constant of integration.

1
Choose the branch.
Take \(a>0\) and \(-1<t<1\), so the substitution covers real \(x\) and \(1-t^2\) is positive.
2
Replace all three quantities.
Use the formulas for \(x\), \(\sqrt{x^2+a^2}\), and \(dx\); do not transform the radical while leaving the differential unchanged.
3
Simplify and integrate.
Cancel factors only after substituting the complete expressions, and integrate the resulting rational function on the parameter interval.
4
Check the result.
Differentiate the proposed antiderivative with respect to \(x\), or verify the transformed derivative and then use \(dt/dx\).

The substitution is not a claim that every integral involving this radical becomes easy. It provides a systematic rationalization; the resulting rational integral may still require additional algebra. Its value is that the radical itself, its branch, and the differential have all been handled explicitly.

Check Your Understanding

Use the branch \(-1<t<1\) and the assumption \(a>0\) when answering.

  1. Derive the expression for \(dx\) by differentiating \(x=2at/(1-t^2)\).
  2. Why does \(\sqrt{x^2+a^2}=a(1+t^2)/(1-t^2)\) have a positive denominator on this branch?
  3. What is the inverse relation expressing \(t\) in terms of \(x\) and \(\sqrt{x^2+a^2}\)?
  4. After substitution, what rational integral represents \(\int dx/\sqrt{x^2+a^2}\)?
  5. Identify the missing factor in the incorrect differential \(dx=2a\,dt/(1-t^2)^2\).

Test yourself: Calculus II final exam practice

Ready to check what you know? Work a full-length practice final, then open the complete worked solutions.

Calculus II Final Exam Practice

Want all 1,114 practice problems?

This tutorial covers one topic. Calc Pass covers the full Calculus AB & BC curriculum (the core of Calculus I and II): full review, every practice question, and step-by-step solutions to each one.

QR code to download the Calc Pass app

Get the Calc Pass App

Scan to download and practice on the go, with instant scoring.

Calc Pass Volume 1: 1,114 Practice Questions book cover

Vol. 1: 1,114 Questions

The complete practice question bank, organized by AB & BC topic.

View on Amazon
Calc Pass Volume 2: Solutions Manual book cover

Vol. 2: Solutions Manual

Fully worked solutions to every question in Volume 1.

View on Amazon

Previous

Next

Back to Tutorials