Introduction
A quadratic expression under a square root can lead to different integration formulas depending on its sign pattern after completing the square. This tutorial focuses on the difference form \(u^2-a^2\), where \(a>0\). Its reciprocal has a logarithmic antiderivative, but only where the radicand is positive.
The domain has two separate parts: \(u>a\) and \(u<-a\). Both are important. On the negative interval, the expression inside the logarithm is negative, so an absolute value is essential for a real-valued antiderivative. When \(u\) is a function of \(x\), the substitution factor and the corresponding intervals for \(x\) must also be handled carefully.
Definition and Domain
Let \(a\) be a positive real number. The expression \(\sqrt{u^2-a^2}\) is real and nonzero precisely when \(u^2-a^2>0\). Factoring the difference gives \((u-a)(u+a)>0\), which holds on the two intervals \(u<-a\) and \(u>a\).
The endpoints \(u=-a\) and \(u=a\) are excluded: there the denominator is zero. There are no real values of the integrand between the endpoints because \(u^2-a^2<0\) for \(-a<u<a\). Thus the formula cannot describe a real antiderivative on one interval that passes through either endpoint.
Why the Logarithm Appears
The expression inside the logarithm is chosen so that its derivative is proportional to the reciprocal square root. Set
On either allowed interval, \(h(u)\ne 0\). For \(u>a\), both terms in \(h(u)\) are positive. For \(u<-a\), the square root is less than \(|u|=-u\), so \(h(u)<0\). In both cases \(\ln|h(u)|\) is defined.
Differentiate on either interval. Since \(a\) is constant,
For a differentiable nonzero function \(h\), the derivative of \(\ln|h|\) is \(h'/h\). This follows by differentiating \(\ln(h)\) where \(h>0\) and \(\ln(-h)\) where \(h<0\); both give \(h'/h\). Therefore,
This verifies the formula and explains why the absolute value works on both branches. Without it, the logarithm would not be real-valued on the negative branch.
Using a Substitution
In applications, \(u\) is often a linear expression in \(x\), such as \(u=mx+b\). If \(m\ne0\), then \(du=m\,dx\), so \(dx=du/m\). When the integrand has exactly the form \(1/\sqrt{u^2-a^2}\) with respect to \(du\), the resulting antiderivative is
This formula is valid on any interval where \(|mx+b|>a\). The factor \(1/m\) comes from the substitution and must not be omitted. The allowed \(x\)-intervals are found by solving \(mx+b>a\) or \(mx+b<-a\); their order depends on the sign of \(m\).
Worked Example: A Difference Form with Two Domain Intervals
Evaluate $$ \int \frac{dx}{\sqrt{(2x-1)^2-9}}. $$
Take \(u=2x-1\), so \(du=2\,dx\) and \(dx=\tfrac12\,du\). Here \(a=3\), and the radicand is positive when \(|u|>3\). Translating the two inequalities gives \(2x-1>3\), or \(x>2\), and \(2x-1<-3\), or \(x<-1\). Thus the integral is real on \((-\infty,-1)\) and \((2,\infty)\).
Substitute and apply the formula: $$ \begin{aligned} \int \frac{dx}{\sqrt{(2x-1)^2-9}} &=\frac12\int\frac{du}{\sqrt{u^2-9}}\\ &=\frac12\ln\left|u+\sqrt{u^2-9}\right|+C\\ &=\frac12\ln\left|2x-1+\sqrt{(2x-1)^2-9}\right|+C. \end{aligned} $$
Check by differentiating. Write \(u=2x-1\); then \(du/dx=2\). The derivative of the proposed antiderivative is $$ \frac12\cdot\frac{1}{\sqrt{u^2-9}}\cdot\frac{du}{dx} = \frac12\cdot\frac{1}{\sqrt{u^2-9}}\cdot 2 = \frac{1}{\sqrt{(2x-1)^2-9}}. $$ This matches the integrand on both allowed intervals.
Returning to the Original Variable
The substitution must preserve the domain information, not just the algebraic form of the integrand. A completed square may produce \(u^2-a^2\) with \(u\) a shifted or scaled version of \(x\). Solve \(|u|>a\) for \(x\) before deciding where the resulting formula is valid. The absolute value in the logarithm remains in place after substituting back.
Worked Example: Completing the Square Before Substitution
Evaluate $$ \int \frac{dx}{\sqrt{4x^2-12x+5}}. $$
Complete the square: $$ \begin{aligned} 4x^2-12x+5 &=4(x^2-3x)+5\\ &=4\left((x-\tfrac32)^2-\tfrac94\right)+5\\ &=4(x-\tfrac32)^2-9+5\\ &=4(x-\tfrac32)^2-4\\ &=(2x-3)^2-2^2. \end{aligned} $$
Set \(u=2x-3\), so \(du=2\,dx\) and \(dx=\tfrac12\,du\). The radicand is positive when \(|u|>2\). The two inequalities give \(2x-3>2\), or \(x>\tfrac52\), and \(2x-3<-2\), or \(x<\tfrac12\). Hence the real domain is \((-\infty,\tfrac12)\cup(\tfrac52,\infty)\).
Now integrate: $$ \begin{aligned} \int \frac{dx}{\sqrt{4x^2-12x+5}} &=\frac12\int\frac{du}{\sqrt{u^2-4}}\\ &=\frac12\ln\left|u+\sqrt{u^2-4}\right|+C\\ &=\frac12\ln\left|2x-3+\sqrt{4x^2-12x+5}\right|+C. \end{aligned} $$
For the check, let \(u=2x-3\), so the square root is \(\sqrt{u^2-4}\). Differentiating the proposed answer gives $$ \frac12\cdot\frac{1}{\sqrt{u^2-4}}\cdot 2 = \frac{1}{\sqrt{u^2-4}} = \frac{1}{\sqrt{4x^2-12x+5}}. $$ The derivative agrees with the integrand throughout each of the two intervals.
Why Absolute Values and Interval Constants Matter
On the branch \(u<-a\), the expression \(u+\sqrt{u^2-a^2}\) is negative. For example, when \(u=-5\) and \(a=3\), it equals \(-5+4=-1\). The logarithm of this quantity is not real, whereas \(\ln|-1|=\ln 1\) is defined. The absolute value is therefore part of the real-valued formula, not a decorative convention.
There is also a domain issue with the constant of integration. An antiderivative is defined on an interval, and the allowed domain here has two disconnected intervals. A formula can be used on both, but its integration constant need not be the same on the left and right. Writing a single \(C\) is common shorthand; more precisely, one may use \(C_1\) on \((-\infty,-a)\) and \(C_2\) on \((a,\infty)\).
Worked Example: A Substitution with Negative Derivative
Evaluate $$ \int \frac{dx}{\sqrt{(5-2x)^2-16}}. $$
Set \(u=5-2x\), so \(du=-2\,dx\) and \(dx=-\tfrac12\,du\). Here \(a=4\). The condition \(|u|>4\) gives \(5-2x>4\), or \(x<\tfrac12\), and \(5-2x<-4\), or \(x>\tfrac92\). Thus the real domain is \((-\infty,\tfrac12)\cup(\tfrac92,\infty)\).
Apply the formula, retaining the negative substitution factor: $$ \begin{aligned} \int \frac{dx}{\sqrt{(5-2x)^2-16}} &=-\frac12\int\frac{du}{\sqrt{u^2-16}}\\ &=-\frac12\ln\left|u+\sqrt{u^2-16}\right|+C\\ &=-\frac12\ln\left|5-2x+\sqrt{(5-2x)^2-16}\right|+C. \end{aligned} $$
To check, let \(u=5-2x\), so \(du/dx=-2\). Differentiation gives $$ -\frac12\cdot\frac{1}{\sqrt{u^2-16}}\cdot(-2) = \frac{1}{\sqrt{(5-2x)^2-16}}. $$ This reproduces the integrand. The two negative factors are both necessary: one comes from the substitution, and the other is the derivative of \(u\) with respect to \(x\).
A Common Mistake
A frequent error is to drop the absolute value because the logarithm is familiar from the positive branch. For \(u<-a\), this would give \(\ln(u+\sqrt{u^2-a^2})\), whose argument is negative and which is not a real-valued function there. The correction is to use \(\ln|u+\sqrt{u^2-a^2}|\) on both intervals.
Another closely related error is to ignore the derivative factor in a substitution. For \(u=5-2x\), writing \(\int dx/\sqrt{u^2-16}=\int du/\sqrt{u^2-16}\) loses the factor \(-\tfrac12\). The correct relation is \(dx=-\tfrac12\,du\), and differentiating the final answer confirms why that factor is required.
Summary
For \(a>0\), the reciprocal square root of \(u^2-a^2\) is real on the two intervals \(u<-a\) and \(u>a\), and on either interval its antiderivative is
When translating back to \(x\), retain the substitution factor, solve the domain condition \(|u|>a\) in terms of \(x\), and keep the absolute value. Treat the two allowed intervals separately when specifying constants of integration.
Check Your Understanding
Use the domain condition and differentiation check as well as the integration formula.
- For which real \(u\) is \(1/\sqrt{u^2-25}\) defined?
- Evaluate \(\int dx/\sqrt{(3x+1)^2-4}\), including its real domain.
- Why is the absolute value necessary in the logarithmic antiderivative when \(u<-a\)?
- For \(u=7-4x\), what factor relates \(dx\) to \(du\)?
- Why may the constants of integration differ on the two domain intervals?
