Why Use This Substitution?
An integral involving \(\sqrt{ax^2+bx+c}\) is generally not a rational integral in \(x\): the square root remains an algebraic expression. When \(a>0\), a particular change of variable turns both \(x\) and the radical into rational functions of a new parameter. This can make an integral rational in that parameter, where standard methods for rational functions apply.
The idea is to choose the parameter as the difference between the radical and the leading linear term. The definition itself is simple; the useful feature is that squaring it gives a linear equation for \(x\).
Deriving the Rational Formulas
Write \(s=\sqrt a\), so \(s>0\), and use \(t=y-sx\). Rearranging gives \(y=sx+t\). Since \(y^2=ax^2+bx+c\), square this rearranged equation and equate it with the radicand. The \(ax^2\) terms cancel, leaving a linear equation for \(x\).
When \(2st-b\ne 0\), we can solve for \(x\), then substitute into \(y=sx+t\). Differentiating the formula for \(x\) gives the change in the differential.
To verify the derivative formula, differentiate \(x=(c-t^2)/(2st-b)\) by the quotient rule:
The numerator is \(-2(st^2-bt+sc)\), which gives the stated expression. Also, the formula for \(y\) follows directly from \(y=sx+t\):
Why the Substitution Rationalizes the Integral
A rational function of \(x\) and \(y\) becomes a rational function of \(t\): the displayed formulas express both \(x\) and \(y\) as quotients of polynomials in \(t\). The differential \(dx\) also contributes a rational factor, \(dx=(dx/dt)\,dt\). Consequently, an integrand that is rational in \(x\) and \(y\), multiplied by \(dx\), transforms into a rational function of \(t\) multiplied by \(dt\).
The method is a change of variable, so it must be used on an interval where the parameter can serve as a local variable. If \(y>0\) and \(2st-b\ne0\), then \(dx/dt=-2y/(2st-b)\ne0\). Thus \(x\) changes locally as \(t\) changes, and the substitution is locally reversible. The formulas do not by themselves authorize crossing a point where their denominator vanishes.
For an actual real-valued integral, the radicand must be nonnegative on the interval in question. The parameter values must also correspond to the chosen principal square root, so the resulting \(y\) must be nonnegative. If a point with \(y=0\) or \(2st-b=0\) separates the interval, work on suitable subintervals or use a different treatment at that point. In the special case \(b^2=4ac\), the radicand is a perfect square and this parameter can be constant on a branch; then the substitution is not invertible there, and the radical should instead be simplified directly.
Set \(t=y-\sqrt a\,x\), where \(y=\sqrt{ax^2+bx+c}\).
Use \(y=\sqrt a\,x+t\), square, cancel \(ax^2\), and isolate \(x\), provided \(2\sqrt a\,t-b\ne0\).
Substitute the formula for \(x\) into \(y=\sqrt a\,x+t\), then differentiate \(x(t)\).
Replace \(x\), \(y\), and \(dx\) in the integrand, then simplify the rational function of \(t\).
After integrating, return to \(x\) using \(t=y-\sqrt a\,x\) and differentiate the proposed antiderivative.
Worked Integrals
Worked Example: Integrating the Reciprocal of a Quadratic Radical
Evaluate \(\int dx/\sqrt{4x^2+9}\). Let \(y=\sqrt{4x^2+9}\). Here \(a=4\), \(b=0\), \(c=9\), and \(\sqrt a=2\). Set \(t=y-2x\). The general formulas give
For real \(x\), \(y=\sqrt{4x^2+9}>|2x|\), so \(t=y-2x>0\); in particular, the displayed denominators do not vanish. Substitute \(y\) and \(dx\) into the integral:
Returning to \(x\), an antiderivative is \(-\tfrac12\ln(\sqrt{4x^2+9}-2x)\). To check it, let \(y=\sqrt{4x^2+9}\); then \(y'=4x/y\), so
This equals the original integrand, as required.
Worked Example: Integrating \(x\) Divided by a Quadratic Radical
Evaluate \(\int x\,dx/\sqrt{4x^2+9}\). Use the same parameter and formulas as in the preceding calculation. The factor \(x/y\) becomes
Now multiply by \(dx/dt\). Including this differential factor is essential:
Integrate term by term:
Since \(t+9/t=(t^2+9)/t=2y\), this is \(y/4+C=\sqrt{4x^2+9}/4+C\). Differentiate to check:
The derivative agrees with the integrand.
Worked Example: A Different Quadratic Radical
Evaluate \(\int dx/\sqrt{2x^2+3x+1}\) on an interval where the radicand is positive and the substitution formulas apply. Put \(s=\sqrt2\), \(y=\sqrt{2x^2+3x+1}\), and \(t=y-sx\). The formulas give
Since \(dx/dt=-2y/(2\sqrt2\,t-3)\), dividing by \(y\) cancels the common factor:
The transformed expression is rational before integration; the logarithm appears because the rational function has a linear denominator. For a check, define \(D=2\sqrt2\,t-3\). The formulas imply \(D=(1-t^2)/x\) when \(x\ne0\), but a direct derivative check can use \(dt/dx=-D/(2y)\), obtained by differentiating \(t=y-\sqrt2\,x\) and using \(y'=(4x+3)/(2y)\). Indeed,
Therefore the derivative of the proposed result is \(-\tfrac1{\sqrt2}(2\sqrt2/D)(-D/(2y))=1/y\), the original integrand.
A Common Mistake: Forgetting the Differential
A substitution changes not only the expressions in the integrand but also the differential. For example, in the second worked example, one might replace \(x/y\) by \((9-t^2)/(2(t^2+9))\) and then integrate that expression with respect to \(t\). That is incorrect: the original integral contains \(dx\), not \(dt\).
The correct calculation includes \(dx=-(t^2+9)/(4t^2)\,dt\):
A reliable check is to write the full transformed integral on one line before simplifying. This makes it less likely that a factor from \(dx/dt\) will be lost.
When the Method Is Useful
The substitution is especially useful when the integrand can be expressed using rational combinations of \(x\) and \(y=\sqrt{ax^2+bx+c}\). Once the substitution is made, these combinations and the differential become rational in \(t\), so integration can proceed using methods for rational functions, including algebraic simplification and partial fractions when appropriate.
The method is not a promise that every resulting integral is simple, nor should it be used across a point where the formulas fail. First check the radicand and the denominator \(2\sqrt a\,t-b\); then transform the entire integrand, including \(dx\); finally verify the answer by differentiation.
Check Your Understanding
Use the definition and formulas to answer these questions. For an integral, state any interval condition needed for the substitution.
- For \(y=\sqrt{5x^2+bx+c}\), what parameter \(t\) defines the positive-leading-coefficient Euler substitution?
- Starting from \(y=\sqrt a\,x+t\), which equation results after squaring and canceling the \(ax^2\) terms?
- Why must a transformed integral include the factor \(dx/dt\)?
- What condition on \(2\sqrt a\,t-b\) is needed to use the displayed formulas for \(x\) and \(y\)?
- For \(y=\sqrt{4x^2+9}\), verify that \(t=y-2x\) is positive for every real \(x\).
