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Rationalizing Substitutions · Tutorial 232 of 998

Euler Substitution for Quadratics with Real Roots

Learn how to rationalize a square root of a quadratic with distinct real roots, integrate in the new variable, and check the result.

Intermediate 10 min read

What You'll Learn

  • Factor a quadratic using its two distinct real roots.
  • Define the signed Euler parameter and derive formulas for the original variable and radical.
  • Convert the differential and representative radical integrals into rational integrals.
  • Track the domain and sign needed for a valid substitution.
  • Check antiderivatives by differentiating them.

Why use a substitution based on the roots?

A square root of a quadratic is not a rational function of \(x\), so an integral involving that radical may not be ready for the methods used for rational functions. When the quadratic has real roots, its factorization gives a substitution that expresses both \(x\) and the radical as rational functions of a new variable. The resulting integral can then be treated as a rational integral.

The key is to build the substitution from the factorization rather than guess a parameter. This also makes the sign and domain conditions visible. Throughout, assume the quadratic has a nonzero leading coefficient and two distinct real roots.

Definition: Let \(Q(x)=ax^2+bx+c\), where \(a\ne 0\), and suppose \(Q\) has distinct real roots \(r_1\) and \(r_2\). Then \(Q(x)=a(x-r_1)(x-r_2)\). On an interval where \(Q(x)>0\) and \(x\ne r_1\), define the Euler parameter \(t\) by \(\sqrt{Q(x)}=t(x-r_1)\), or equivalently \(t=\frac{\sqrt{Q(x)}}{x-r_1}\). The parameter is signed: it is positive or negative according to the sign of \(x-r_1\).

The substitution is applied on intervals where the radical is real and \(x\ne r_1\). If \(a>0\), \(Q(x)>0\) outside the two roots; if \(a<0\), it is positive between them. A repeated root is excluded: in that case the relation does not give the same useful parametrization, and the radical can instead be handled directly using the repeated factor.

Deriving the rational formulas

Squaring the defining equation gives \(Q(x)=t^2(x-r_1)^2\). Use the factorization, then divide by \(x-r_1\), which is nonzero on the interval under consideration:

$$ a(x-r_1)(x-r_2)=t^2(x-r_1)^2 \quad\Longrightarrow\quad a(x-r_2)=t^2(x-r_1). $$

Collect the terms involving \(x\) on one side. Since the roots are distinct, \(a-t^2\) cannot be zero at a finite \(x\) satisfying this equation: if \(t^2=a\), it would imply \(a(x-r_2)=a(x-r_1)\), hence \(r_1=r_2\). Therefore, we can solve for \(x\). Subtracting \(r_1\) from the resulting expression gives a convenient formula for the radical as well:

$$ (a-t^2)x=ar_2-t^2r_1, \qquad x=\frac{ar_2-t^2r_1}{a-t^2}, \qquad x-r_1=\frac{a(r_2-r_1)}{a-t^2}. $$

Now use the original, unsquared substitution \(\sqrt{Q(x)}=t(x-r_1)\), and differentiate the expression for \(x\):

$$ \sqrt{Q(x)} =\frac{at(r_2-r_1)}{a-t^2}, \qquad \frac{dx}{dt} =\frac{2at(r_2-r_1)}{(a-t^2)^2}. $$

These expressions show why the substitution works. Both \(x\) and \(\sqrt{Q(x)}\) are rational functions of \(t\), as is \(dx/dt\). Consequently, an integrand built from rational operations on \(x\) and \(\sqrt{Q(x)}\), multiplied by \(dx\), becomes a rational function of \(t\) times \(dt\), wherever its denominators are nonzero.

Important condition: The equation for \(x\) was obtained by dividing by \(x-r_1\). The formulas therefore describe the substitution away from \(r_1\). If an integral reaches that root, work on an interval not containing the root and treat any endpoint by the appropriate limit.

Worked Example: Integrating a radical with two exterior branches

Worked Example: \(\int \frac{dx}{\sqrt{x^2-1}}\) for \(x>1\)

Factor the radicand: \(x^2-1=(x+1)(x-1)\). Thus \(a=1\), \(r_1=-1\), and \(r_2=1\). Use \(\sqrt{x^2-1}=t(x+1)\). On \(x>1\), both the radical and \(x+1\) are positive, so \(t>0\).

The general formulas give

$$ x=\frac{1+t^2}{1-t^2}, \qquad \sqrt{x^2-1}=\frac{2t}{1-t^2}, \qquad dx=\frac{4t}{(1-t^2)^2}\,dt. $$

Substitute into the integral and cancel the common factors:

$$ \int\frac{dx}{\sqrt{x^2-1}} = \int \frac{\frac{4t}{(1-t^2)^2}}{\frac{2t}{1-t^2}}\,dt = \int\frac{2}{1-t^2}\,dt. $$

Since \(\frac{2}{1-t^2}=\frac{1}{1-t}+\frac{1}{1+t}\), integration yields \(-\ln|1-t|+\ln|1+t|+C=\ln\left|\frac{1+t}{1-t}\right|+C\). For this branch, \(0<t<1\), so the absolute value can be omitted. Also,

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

Thus an antiderivative is \(\ln(x+\sqrt{x^2-1})+C\). To check, differentiate it:

$$ \frac{d}{dx}\ln(x+\sqrt{x^2-1}) = \frac{1+\frac{x}{\sqrt{x^2-1}}}{x+\sqrt{x^2-1}} = \frac{1}{\sqrt{x^2-1}}. $$

The derivative agrees with the integrand on \(x>1\).

Worked Example: A quadratic that is positive between its roots

Worked Example: \(\int \frac{dx}{\sqrt{-x^2+4x-3}}\) for \(1<x<3\)

Factor the radicand as \(-x^2+4x-3=-(x-1)(x-3)\). Here \(a=-1\), \(r_1=1\), and \(r_2=3\). On \(1<x<3\), the radicand and \(x-r_1\) are positive, so \(t=\frac{\sqrt{-x^2+4x-3}}{x-1}>0\).

Substitution into the formulas gives

$$ x=\frac{t^2+3}{t^2+1}, \qquad x-1=\frac{2}{t^2+1}, \qquad \sqrt{-x^2+4x-3}=\frac{2t}{t^2+1}. $$

Differentiating \(x\) gives

$$ \frac{dx}{dt} = \frac{2t(t^2+1)-2t(t^2+3)}{(t^2+1)^2} = -\frac{4t}{(t^2+1)^2}. $$

Therefore, the integral becomes

$$ \int\frac{dx}{\sqrt{-x^2+4x-3}} = \int \frac{-\frac{4t}{(t^2+1)^2}}{\frac{2t}{t^2+1}}\,dt = \int-\frac{2}{1+t^2}\,dt = -2\arctan t+C. $$

Returning to \(x\), this is \(-2\arctan\left(\frac{\sqrt{-x^2+4x-3}}{x-1}\right)+C\). To check it, set \(F(x)=-2\arctan t\), where \(t^2=\frac{3-x}{x-1}\). Differentiating this relation gives \(2tt'=-\frac{2}{(x-1)^2}\), so \(t'=-\frac{1}{t(x-1)^2}\). Also \(1+t^2=\frac{2}{x-1}\). Hence

$$ F'(x) = -\frac{2t'}{1+t^2} = \frac{2}{t(x-1)^2}\cdot\frac{x-1}{2} = \frac{1}{t(x-1)} = \frac{1}{\sqrt{-x^2+4x-3}}. $$

This verifies the antiderivative throughout \(1<x<3\).

Worked Example: A rational factor in the integrand

Worked Example: \(\int \frac{dx}{x\sqrt{x^2-1}}\) for \(x>1\)

Use \(x^2-1=(x+1)(x-1)\) and the same choice \(r_1=-1\), \(r_2=1\). The formulas from the first example are \(x=\frac{1+t^2}{1-t^2}\), \(\sqrt{x^2-1}=\frac{2t}{1-t^2}\), and \(dx=\frac{4t}{(1-t^2)^2}\,dt\). Substitution gives

$$ \int\frac{dx}{x\sqrt{x^2-1}} = \int \frac{\frac{4t}{(1-t^2)^2}} {\frac{1+t^2}{1-t^2}\cdot\frac{2t}{1-t^2}}\,dt = \int\frac{2}{1+t^2}\,dt = 2\arctan t+C. $$

Thus the answer in \(x\) is \(2\arctan\left(\frac{\sqrt{x^2-1}}{x+1}\right)+C\). For a direct check, \(t^2=\frac{x-1}{x+1}\), so \(2tt'=\frac{2}{(x+1)^2}\) and \(t'=\frac{1}{t(x+1)^2}\). Since \(1+t^2=\frac{2x}{x+1}\),

$$ \frac{d}{dx}(2\arctan t) = \frac{2t'}{1+t^2} = \frac{2}{t(x+1)^2}\cdot\frac{x+1}{2x} = \frac{1}{xt(x+1)} = \frac{1}{x\sqrt{x^2-1}}. $$

The last equality uses \(t(x+1)=\sqrt{x^2-1}\), which follows from the chosen substitution.

Why the sign matters

The equation \(\sqrt{Q(x)}=t(x-r_1)\) is not interchangeable with its squared version unless the sign is also respected. Squaring is useful for deriving the rational formulas, but it cannot tell us which sign of \(t\) matches the original radical. The signed definition \(t=\frac{\sqrt{Q(x)}}{x-r_1}\) resolves that issue.

Common mistake: For \(Q(x)=x^2-1\) at \(x=-2\), the radical is \(\sqrt{3}\), while \(x+1=-1\). The signed parameter is \(t=\frac{\sqrt{3}}{-1}=-\sqrt{3}\), and \(t(x+1)=\sqrt{3}\), as required. If one incorrectly insists that \(t\) must be positive, then \(t=\sqrt{3}\) gives \(t(x+1)=-\sqrt{3}\), not the positive square root. The correction is to use the signed ratio and keep track of the interval on which the substitution is made.

Using the method effectively

For an integral with a quadratic radical, first check that the quadratic has two distinct real roots and factor it as \(a(x-r_1)(x-r_2)\). Choose one root as \(r_1\), define \(t\) from the unsquared equation, and derive \(x\), \(\sqrt{Q(x)}\), and \(dx\) in terms of \(t\). Then substitute all three into the integrand. Once the result is rational in \(t\), apply the appropriate rational-integration method.

1
Factor the quadratic.
Identify \(a\), \(r_1\), and \(r_2\), and determine where the radicand is positive.
2
Define the signed parameter.
Use \(t=\frac{\sqrt{Q(x)}}{x-r_1}\) on an interval where \(Q(x)>0\) and \(x\ne r_1\).
3
Rationalize and integrate.
Substitute the formulas for \(x\), the radical, and \(dx\), then integrate the resulting rational function.
4
Return and check.
Write the result in \(x\) and differentiate it on the interval where it was derived.

The substitution is a way to convert one class of algebraic integrals into rational ones; it does not by itself guarantee that the final rational integral is easy. Its value is that it removes the square root while preserving enough information to reverse the substitution and verify the result.

Check Your Understanding

Use the factorization and the signed parameter carefully when answering these questions.

  1. For \(Q(x)=2(x-1)(x-4)\), write the Euler substitution using \(r_1=1\), then find the formulas for \(x\) and \(\sqrt{Q(x)}\) in terms of \(t\).
  2. Why must the roots be distinct in the derivation that solves for \(x\) using the denominator \(a-t^2\)?
  3. For a quadratic with negative leading coefficient and distinct real roots, where is its square root real and positive?
  4. What sign does \(t\) have when \(x<r_1\) and \(Q(x)>0\)? Explain using its definition.
  5. After substituting in an integral, what feature of the resulting expression indicates that the square root has been rationalized?

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