Why the Choice Matters
A square root of a quadratic can make an integral look algebraic rather than rational. An Euler substitution introduces a parameter \(t\) that expresses both \(x\) and the square root as rational functions of \(t\). The original integral can then be treated as a rational integral, provided the transformed expressions are defined on the interval in question.
There are three standard choices. One uses a positive leading coefficient, another a positive constant term, and the third a real root of the quadratic. These choices are not mutually exclusive: more than one may apply to the same quadratic. The practical question is which choice gives manageable algebra on the interval of integration.
The Three Choices and Their Conditions
The leading-coefficient substitution is available when \(a>0\). Set \(y=\sqrt a\,x+t\). The constant-term substitution is available when \(c>0\). Set \(y=tx+\sqrt c\). The real-root substitution is available when \(q\) has a simple real root \(r\); set \(y=t(x-r)\). In each case, the equation \(y^2=q(x)\) determines \(x\) as a rational function of \(t\).
A root is simple when \(q'(r)=2ar+b\ne 0\). This condition matters: a double root does not give the usual inverse formula for the real-root substitution. The formulas below also require their denominators to be nonzero. If a denominator vanishes at an isolated parameter value, the substitution may still be used on an interval that avoids that value.
Deriving the Rational Forms
The algebra behind each choice is short, but keeping the signs consistent is essential. In every case, substitute the proposed relation into \(y^2=q(x)\), solve for \(x\), and then use the substitution itself to express \(y\).
With \(a>0\) and \(y=\sqrt a\,x+t\), expand \(y^2-ax^2\) and equate it to \(bx+c\).
With \(c>0\) and \(y=tx+\sqrt c\), expand \(y^2-c\) and equate it to \(ax^2+bx\).
With \(q(r)=0\) and \(y=t(x-r)\), factor \(q(x)\) around \(r\), then solve after dividing by \(x-r\) on an interval where \(x\ne r\).
For the first choice, expansion gives \(2\sqrt a\,xt+t^2=bx+c\). For the second, it gives \(t^2x^2+2\sqrt c\,tx=ax^2+bx\). For the third, expansion around the root gives \(q(x)=a(x-r)^2+(2ar+b)(x-r)\). Solving these equations yields the following formulas.
For example, the last formula follows from \(t^2(x-r)^2=a(x-r)^2+(2ar+b)(x-r)\). Since the substitution excludes \(x=r\), divide by \(x-r\), giving \(t^2(x-r)=a(x-r)+(2ar+b)\). Rearranging gives \((t^2-a)(x-r)=2ar+b\), as claimed. This derivation also shows why a simple root is needed: if \(2ar+b=0\), it does not produce an inverse formula for \(x\).
Worked Example: Positive Leading Coefficient
Worked Example: Positive Leading Coefficient
Use the leading-coefficient substitution to evaluate \(\displaystyle\int\frac{dx}{\sqrt{4x^2+1}}\). Here \(a=4>0\), so set \(y=\sqrt{4x^2+1}=2x+t\). Then \(t=y-2x\). From \(y^2-4x^2=1\),
Thus \(4xt=1-t^2\), so \(x=(1-t^2)/(4t)\). Substitution into \(y=2x+t\) gives
Differentiate \(x=(1-t^2)/(4t)=1/(4t)-t/4\):
Therefore an antiderivative is \(-\tfrac12\ln t\), where \(t=\sqrt{4x^2+1}-2x>0\). To check, differentiate \(t\): since \(y=\sqrt{4x^2+1}\), \(dt/dx=4x/y-2=-2t/y\). Hence
Worked Example: Positive Constant Term
Worked Example: Positive Constant Term
Evaluate \(\displaystyle\int\frac{dx}{\sqrt{x^2+4}}\) using the constant-term substitution. Since \(c=4>0\), set \(y=\sqrt{x^2+4}=tx+2\). Squaring and subtracting \(4\) gives
For \(x\ne0\), divide by \(x\) to obtain \(x(t^2-1)+4t=0\), so \(x=4t/(1-t^2)\). The resulting expression also extends to \(x=0\) at \(t=0\). Now
Differentiate the expression for \(x\):
Integrating gives \(\ln\!\left(\frac{1+t}{1-t}\right)+C\), with \(-1<t<1\) for real \(x\). The check is direct: the derivative of this logarithm with respect to \(t\) is \(1/(1+t)+1/(1-t)=2/(1-t^2)= (dx/dt)/y\). Thus, by the chain rule, its derivative with respect to \(x\) is \(1/y=1/\sqrt{x^2+4}\).
Worked Example: A Simple Real Root
Worked Example: A Simple Real Root
Evaluate \(\displaystyle\int\frac{dx}{\sqrt{x^2-1}}\) on \(x>1\). The quadratic has the simple root \(r=1\), since \(q'(1)=2\ne0\). Set \(y=\sqrt{x^2-1}=t(x-1)\). Because \(x>1\), this gives \(t>0\). Using \(x^2-1=(x-1)(x+1)\), or the general real-root formula, gives
For \(x>1\), the corresponding parameter satisfies \(t>1\), so the denominators are nonzero. Differentiating \(x=1+2/(t^2-1)\) gives
An antiderivative in \(t\) is \(\ln\left|\frac{t+1}{t-1}\right|\), because its derivative is \(1/(t+1)-1/(t-1)=-2/(t^2-1)\). By the chain rule and the computed \(dx/y\), this expression has derivative \(1/y=1/\sqrt{x^2-1}\) with respect to \(x\). This also checks the transformed integral.
Comparing the Choices
The examples use the same type of integrand, \(dx/y\), so their transformed algebra can be compared directly. The choice of substitution determines the parameter and the rational expressions for \(x\), \(y\), and \(dx/y\).
| Quadratic and choice | Parameter relation | Transformed \(dx/y\) |
|---|---|---|
| \(4x^2+1\), leading coefficient | \(t=y-2x\) | \(-dt/(2t)\) |
| \(x^2+4\), constant term | \(t=(y-2)/x\) | \(2\,dt/(1-t^2)\) |
| \(x^2-1\), root \(r=1\) | \(t=y/(x-1)\) | \(-2\,dt/(t^2-1)\) |
These are illustrations, not a ranking. The first gives a logarithmic differential immediately; the other two give rational functions with quadratic denominators. For a different integrand, the simplest choice may differ. In general, if the original integrand is a rational expression in \(x\) and \(y\), and the chosen substitution expresses both rationally in \(t\), then \(dx\) is rational in \(t\) as well. The transformed integral is therefore rational in \(t\), wherever the substitution is defined.
A Sign Error to Avoid
A common mistake is to reverse the terms when solving for \(x\). For \(y=\sqrt{4x^2+1}=2x+t\), the correct equation is \(t(t+4x)=1\). It gives \(4xt=1-t^2\), hence \(x=(1-t^2)/(4t)\). Reversing the numerator and writing \(x=(t^2-1)/(4t)\) is incorrect.
Indeed, with that incorrect value, \(t(t+4x)=t\bigl(t+(t^2-1)/t\bigr)=2t^2-1\), which is not identically \(1\). The proposed \(x\) therefore fails the equation \(y^2-4x^2=1\). A reliable check is to substitute the solved expression back into the equation that produced it before differentiating.
How to Make the Choice
Inspect \(a\), \(c\), and the roots before introducing a parameter. If \(a>0\), the leading-coefficient form is available; if \(c>0\), the constant-term form is available; if there is a simple real root, the real-root form is available. Then check the interval: the radicand must be nonnegative, the chosen parameter must represent the required square-root branch, and the inverse formula must have a nonzero denominator there.
When several choices apply, compare the resulting \(x(t)\), \(y(t)\), and \(dx/dt\) rather than choosing by a rule that claims one form is always best. A useful substitution makes the algebra shorter and stays valid across the interval under consideration. Checking the defining equation and differentiating the final antiderivative provide independent checks on the calculation.
Check Your Understanding
For each question, identify the applicable substitution or explain the condition it requires.
- For \(q(x)=3x^2-2x+5\), which of the three Euler substitutions are available by their coefficient or root conditions?
- For \(q(x)=x^2-4x+3\), choose a simple real root and write the real-root substitution.
- Why does a double root not yield the usual inverse formula for the real-root substitution?
- For the leading-coefficient substitution, why must the sign in \(x=(c-t^2)/(2\sqrt a\,t-b)\) be checked against the defining equation?
- What must be true of the original integrand for an Euler substitution to make it a rational function of the parameter?
