Why Simplify Before Integrating?
For a differentiable graph \(y=f(x)\) on an interval \([a,b]\), the arc-length formula is \(L=\int_a^b\sqrt{1+[f'(x)]^2}\,dx\), provided \(f'\) is continuous on that interval. The formula gives the length, but its integrand is not always in a useful form at first glance.
Before trying to integrate, inspect the expression inside the square root. Writing the derivative as a fraction, using a common denominator, or applying an algebraic identity can turn a complicated-looking radical into a simpler function. Sometimes the resulting integrand has an elementary antiderivative; sometimes the rewrite still leaves an integral that needs another method or cannot be expressed with elementary functions.
A Reliable Algebraic Pattern
Suppose the derivative can be written as \(f'(x)=P(x)/Q(x)\), where \(Q(x)\ne 0\) throughout the interval. Then
The absolute value appears because \(\sqrt{Q(x)^2}=|Q(x)|\), not \(Q(x)\) in general. If the expression \(Q(x)^2+P(x)^2\) factors as \(R(x)^2\), the integrand becomes \(|R(x)|/|Q(x)|\). On an interval where the signs of \(R\) and \(Q\) are known, those absolute values can be removed correctly.
A useful identity for spotting perfect squares is
Indeed, expanding the left-hand side gives \(u^2-2uv+v^2+4uv=u^2+2uv+v^2\), which is \((u+v)^2\). This identity often appears when the numerator and denominator of a derivative have related algebraic forms.
Worked Example: A Constant Slope
Worked Example: A Constant Slope
Find the length of \(f(x)=\frac{3}{4}x\) from \(x=0\) to \(x=2\).
Differentiate the graph:
Substitute into the length integrand and simplify the constant under the radical:
Therefore the length is
Check the evaluation by differentiating the antiderivative: \(\frac{d}{dx}(5x/4)=5/4\), the simplified integrand. Thus the length is \(5/2\).
This example is simple, but it illustrates the first check: evaluate \(1+[f'(x)]^2\) exactly before considering any more involved integration. A constant positive quantity under the square root gives a constant integrand.
Derivation: When a Rational Derivative Produces a Square
The common-denominator rewrite is useful because it reduces the simplification question to an algebraic one. The following derivation states exactly what happens when the numerator is a perfect square.
Assume \(f'(x)=P(x)/Q(x)\) on an interval \(I\), with \(Q(x)\ne 0\) for every \(x\in I\).
Then \(1+[f'(x)]^2=(Q(x)^2+P(x)^2)/Q(x)^2\).
If \(Q(x)^2+P(x)^2=R(x)^2\), then \(\sqrt{1+[f'(x)]^2}=|R(x)|/|Q(x)|\), because the square root of a square is an absolute value.
If \(R\) and \(Q\) have constant, known signs on \(I\), simplify the absolute values using those signs. If either sign changes, retain the absolute value or split the interval at the sign change.
This derivation is also a practical test. A common denominator alone does not guarantee a simpler integral. The main opportunity occurs when \(Q^2+P^2\) factors as a square, or otherwise reduces to a recognizable expression.
Worked Example: A Rational Derivative and a Perfect Square
Worked Example: A Rational Derivative and a Perfect Square
Find the length of \(f(x)=-\ln(1-x^2)\) on \([0,\frac12]\).
Since \(1-x^2>0\) on this interval, the function is defined there. Differentiating gives
Use a common denominator in the expression under the square root:
Expand and factor the numerator:
On \([0,\frac12]\), \(1+x^2>0\) and \(1-x^2>0\). Hence the square root simplifies without a sign change:
Now integrate the simplified expression. Since \(\frac{d}{dx}\ln\left(\frac{1+x}{1-x}\right)=\frac{2}{1-x^2}\) on this interval, an antiderivative is \(-x+\ln\left(\frac{1+x}{1-x}\right)\). Therefore
Check by differentiation: the derivative of the stated antiderivative is \(-1+2/(1-x^2)\), exactly the simplified length integrand. The endpoint value is positive, as a length must be.
Worked Example: A Radical Derivative
Worked Example: A Radical Derivative
Find the length of \(f(x)=-\sqrt{1-x^2}\) on \([0,\frac12]\).
Differentiate, noting that \(1-x^2>0\) on the interval:
Combine the terms under the square root using the denominator \(1-x^2\):
Thus \(\sqrt{1+[f'(x)]^2}=1/\sqrt{1-x^2}\), since the denominator is positive on the interval. The length is
Check by differentiating: \(\frac{d}{dx}\arcsin x=1/\sqrt{1-x^2}\) for \(-1<x<1\), which is precisely the simplified integrand. The interval lies within that domain.
When Does Simplification Lead to an Exact Evaluation?
Algebraic simplification and integration are separate decisions. A simplified integrand is especially useful when it becomes a constant, a rational function with a known antiderivative, or a standard function such as \(1/\sqrt{1-x^2}\). The examples above show each type of outcome: a constant, a rational expression, and a standard inverse-trigonometric derivative.
By contrast, if the numerator \(Q^2+P^2\) does not factor usefully, simplifying the expression may not make the integral easier. Do not force a perfect-square pattern that is not present. Keep the exact expression, check whether a different algebraic form is recognizable, and only then decide how to proceed. An exact setup does not by itself imply an elementary antiderivative.
| What you see after rewriting | Useful next check |
|---|---|
| A constant under the square root | Take its exact square root and integrate the constant. |
| A quotient with a perfect-square numerator | Take the square root carefully, keeping absolute values. |
| A familiar standard integrand | Check its domain and use a matching antiderivative. |
| No useful factorization or recognizable form | Do not assume the integral has an elementary antiderivative. |
Common Pitfall: Dropping an Absolute Value
A frequent error is to use \(\sqrt{A(x)^2}=A(x)\) without checking the sign of \(A(x)\). In fact, \(\sqrt{A(x)^2}=|A(x)|\). For example, take \(f'(x)=(x^2-1)/(2x)\) on \([-2,-1]\). Algebra gives
The incorrect simplification is \((x^2+1)/(2x)\), obtained by dropping the absolute value in \(\sqrt{4x^2}=2|x|\). Since \(x<0\) on the whole interval, this incorrect expression is negative. At \(x=-1\), it equals \(-1\), even though a square root cannot be negative.
The correct simplification is
Here \(x^2+1>0\) and \(|x|>0\), so the result is positive throughout the interval, as the original square root requires. Check signs on the actual interval, not just at a convenient positive value.
Why the Algebra Matters
The arc-length integrand is always nonnegative, but its original form can conceal the structure that makes it integrable. Factoring and finding common denominators reveal whether the terms combine into a square or a familiar expression. This can turn an apparently difficult radical into an ordinary integral while preserving the exact length.
The most dependable sequence is: compute \(f'(x)\), rewrite \(1+[f'(x)]^2\) exactly, factor where possible, account for absolute values, and then choose an integration method for the simplified integrand. Each step has a clear purpose, and the sign check prevents an algebraic simplification from changing the meaning of the square root.
Check Your Understanding
For each question, focus first on the algebra inside the square root and the signs on the stated interval.
- If \(f'(x)=P(x)/Q(x)\), with \(Q(x)\ne0\), what expression results after combining \(1+[f'(x)]^2\) over a common denominator?
- Why is \(\sqrt{Q(x)^2}=|Q(x)|\), rather than always \(Q(x)\)?
- For \(f'(x)=2x/(1-x^2)\) on \([0,\frac12]\), what perfect square appears in the numerator after combining terms?
- What sign check is needed before removing absolute values from \(|R(x)|/|Q(x)|\)?
- Does an algebraically simplified length integrand always have an elementary antiderivative? Explain briefly.