Why Use a Ratio of the Linear Factors?
When an integrand contains the square root of a product of two linear factors, a substitution based on their ratio can remove the radical. The substitution is useful because both the original variable and the square root can be written as rational functions of the new parameter. The signs and parameter intervals matter: the same algebraic formula for the inverse substitution covers two distinct real branches.
Assume the two distinct real roots are ordered as \(r<s\). We consider the principal square root, which is nonnegative wherever its radicand is nonnegative. The product \((x-r)(x-s)\) is nonnegative when \(x\le r\) or \(x\ge s\), and is negative for \(r<x<s\). The substitution is therefore considered separately on the two open intervals \(x<r\) and \(x>s\); at the endpoints the square root is zero and the substitution formulas have boundary limits.
Solving for the Variable and the Radical
Squaring the definition gives \(t^2=(x-r)/(x-s)\). Solving this equation for \(x\) gives an inverse formula, valid when \(t\ne1\). Substitution into each linear factor then gives the radical. These algebraic steps determine the formulas; the absolute value in the square root must still be evaluated according to the branch.
Multiplying the last two expressions gives \((x-r)(x-s)=(s-r)^2t^2/(t^2-1)^2\). Since \(s-r>0\) and \(t>0\), taking the principal square root produces an absolute value in the denominator.
Differentiate the inverse formula to obtain the differential needed in an integral:
The derivative is negative on both parameter intervals. Thus, as \(x\) increases along either branch, \(t\) decreases. On the left branch, \(x\to-\infty\) corresponds to \(t\to1^-\), while \(x\to r^-\) corresponds to \(t\to0^+\). On the right branch, \(x\to s^+\) corresponds to \(t\to+\infty\), while \(x\to+\infty\) corresponds to \(t\to1^+\). The point \(t=1\) is not attained at any finite \(x\).
Why the Substitution Rationalizes the Integrand
The method works because the inverse formula expresses \(x\) as a rational function of \(t\); the square root is also rational in \(t\) after its branch is fixed. The derivative \(dx/dt\) is rational as well. Therefore, whenever an integrand is a rational expression in \(x\) and \(\sqrt{(x-r)(x-s)}\), substituting these formulas produces a rational expression in \(t\), wherever the original integrand and the substitution are defined.
The substitution is reversible on each branch, not across both branches at once. Indeed, the map \(x\mapsto t\) takes \(x<r\) onto \(0<t<1\), and \(x>s\) onto \(t>1\). The inverse formula sends those parameter intervals back to their respective \(x\)-intervals. This interval information also determines the sign in the radical formula and the limits in definite integrals.
Worked Examples
Worked Example: Integrating on the Right-Hand Branch
Evaluate an antiderivative of \(1/\sqrt{(x-1)(x-4)}\) on \(x>4\). Here \(r=1\), \(s=4\), so \(t=\sqrt{(x-1)/(x-4)}>1\). The formulas give \(x=(4t^2-1)/(t^2-1)\), \(\sqrt{(x-1)(x-4)}=3t/(t^2-1)\), and \(dx/dt=-6t/(t^2-1)^2\). Thus
Since \(-2/(t^2-1)=-1/(t-1)+1/(t+1)\), integration gives \(-\ln|t-1|+\ln|t+1|+C\). On \(t>1\), both \(t-1\) and \(t+1\) are positive, so the result is \(\ln((t+1)/(t-1))+C\). Returning to \(x\), one may substitute \(t=\sqrt{(x-1)/(x-4)}\).
To check, differentiate with respect to \(t\): \(d[\ln((t+1)/(t-1))]/dt=1/(t+1)-1/(t-1)=-2/(t^2-1)\), which is the transformed integrand. Since \(dt/dx\) is the reciprocal of \(dx/dt\) on this branch, the chain rule gives derivative \(1/\sqrt{(x-1)(x-4)}\) with respect to \(x\), as required.
Worked Example: A Definite Integral on the Left-Hand Branch
Evaluate \(\int_{-3}^{0} dx/\sqrt{(x-1)(x-4)}\). Both endpoints lie to the left of \(r=1\), so \(0<t<1\). Using \(t=\sqrt{(x-1)/(x-4)}\), the transformed radical is \(3t/(1-t^2)\), while \(dx/dt=-6t/(1-t^2)^2\). Hence the transformed integrand is
An antiderivative with respect to \(t\) is \(\ln((1-t)/(1+t))\), because its derivative is \(-1/(1-t)-1/(1+t)=-2/(1-t^2)\). The endpoint parameters are \(t(-3)=\sqrt{4/7}=2/\sqrt7\) and \(t(0)=\sqrt{1/4}=1/2\). Therefore the definite integral is
The logarithm’s argument is positive because \(\sqrt7-2>0\). It is greater than \(1\): the equivalent inequality \(\sqrt7+2>3(\sqrt7-2)\) reduces to \(8>2\sqrt7\), which holds since \(4>\sqrt7\). Rationalizing the ratio gives \((11+4\sqrt7)/9\), since \((\sqrt7+2)^2=11+4\sqrt7\) and \(3((\sqrt7)^2-2^2)=9\). This is also greater than \(1\), as \(11+4\sqrt7>9\). Differentiating the chosen \(t\)-antiderivative gives the transformed integrand, and the endpoint substitution above checks the definite evaluation.
Worked Example: A Rational Function of the Parameter
For \(x>5\), find an antiderivative of \(1/((x-2)\sqrt{(x-2)(x-5)})\). Take \(r=2\), \(s=5\), so \(t=\sqrt{(x-2)/(x-5)}>1\). Then
Substituting all three expressions gives
Since \(1/t=\sqrt{(x-5)/(x-2)}\), an antiderivative is \((2/3)\sqrt{(x-5)/(x-2)}+C\). To check directly, the derivative of \((x-5)/(x-2)\) is \(3/(x-2)^2\). The chain rule therefore gives
where the last equality uses \(x>5\), so \(x-2\) and \(x-5\) are positive. The derivative matches the original integrand.
A Branch-Sign Pitfall
A frequent error is to remove the absolute value from the radical formula without checking which side of \(t=1\) applies. For \(r=1\), \(s=4\), and \(x=0\), the parameter is \(t=1/2\). The actual radical is \(\sqrt{(-1)(-4)}=2\). The incorrect use of \(3t/(t^2-1)\) gives \(3(1/2)/(1/4-1)=-2\), which cannot equal a principal square root. On this left branch the correct formula is \(3t/(1-t^2)=2\). On the right branch \(t>1\), the correct denominator is instead \(t^2-1\).
When This Method Is Useful
This substitution is especially effective when both the variable and the square root appear in rational combinations, because it converts those combinations to rational functions of \(t\). It does not make every resulting integral immediate: the rational function may still require algebraic simplification or a suitable integration technique. The essential achievement is that the radical has been removed without losing control of the real domain.
For real integration, do not use the formula on \(r<x<s\), where \((x-r)(x-s)<0\) and the square root is not real. If the roots are initially named in the opposite order, first relabel them so that \(r<s\). At \(x=r\) or \(x=s\), the ratio substitution may have a zero or an unbounded parameter; handle such endpoints by limits when evaluating an improper integral, rather than treating the inverse formula as valid at the endpoint.
Check Your Understanding
Use the branch conditions as well as the algebra when answering these questions.
- For \(r<s\), on which real intervals is \(\sqrt{(x-r)(x-s)}\) defined?
- Derive \(x=(st^2-r)/(t^2-1)\) from \(t^2=(x-r)/(x-s)\).
- What are the ranges of \(t\) when \(x<r\) and when \(x>s\)?
- Why does the radical formula contain \(|t^2-1|\) before a branch is chosen?
- For \(r=1\), \(s=4\), and \(x=0\), compute \(t\) and verify the correct radical formula gives the principal square root.
