Why Compare Antiderivative Forms?
One integral can lead to formulas that look quite different. Algebraic rearrangements may hide that two expressions differ by a constant, while logarithm identities can make distinct-looking answers exactly equal. Comparing such forms is useful because it checks whether a result is correct without relying on how it was obtained.
The reliable test is differentiation. If two differentiable functions have the same derivative throughout an interval, their difference is constant there. The interval condition matters: on a disconnected domain, different components can have different constants.
A Basic Algebraic Comparison
Worked Example: Two Polynomial Forms
Compare \(F(x)=x^2+2x\) and \(G(x)=(x+1)^2\). First differentiate each function:
The derivatives agree for every real \(x\), so the functions are antiderivatives of the same function. Expand \(G\) to find their difference:
Thus \(G=F+1\) on \(\mathbb R\). As a check, differentiating either formula gives \(2x+2\), as required. The additive constant does not change the derivative.
This example illustrates the main idea: the formulas need not be identical to be valid answers for the same antiderivative problem. Their derivatives determine whether they represent the same integrand; subtracting them identifies the constant difference.
The Zero-Derivative Principle
The comparison rests on a fundamental fact from differential calculus. Suppose \(F\) and \(G\) are differentiable on an interval \(I\), and \(F'(x)=G'(x)\) for every \(x\in I\). Set \(H=F-G\). Then \(H'(x)=0\) on \(I\). The Mean Value Theorem shows that \(H\) has the same value at any two points of the interval.
Define \(H=F-G\). The assumption that \(F'=G'\) gives \(H'=F'-G'=0\).
Let \(a,b\in I\) with \(a<b\). Since \(I\) is an interval, the whole segment \([a,b]\) lies in \(I\).
\(H\) is continuous on \([a,b]\) and differentiable on \((a,b)\). Therefore some \(c\in(a,b)\) satisfies \(H(b)-H(a)=H'(c)(b-a)=0\). Hence \(H(a)=H(b)\).
Since the two points were arbitrary, \(H\) has the same value everywhere on \(I\). Thus \(F-G=C\). Conversely, differentiating \(F-G=C\) gives \(F'-G'=0\), so \(F'=G'\).
The interval hypothesis connects every pair of points by a segment on which the Mean Value Theorem applies. If the domain has separate pieces, this argument establishes constancy on each interval component, not necessarily the same constant across all components.
Logarithmic Forms and Additive Constants
Logarithmic antiderivatives often differ by constants because multiplying the argument of a logarithm by a positive constant adds a constant to its value. Absolute values make the standard logarithmic derivative formula valid on intervals that may lie on either side of zero: for \(x\ne 0\), the derivative of \(\ln|x|\) is \(1/x\).
Worked Example: A Scaled Logarithm
On either interval \((-\infty,0)\) or \((0,\infty)\), compare \(F(x)=\ln|x|\) and \(G(x)=\ln|3x|\). Differentiate both:
The derivatives agree on each interval. The logarithm identity gives the difference explicitly:
Therefore the two antiderivatives differ by the constant \(\ln 3\). Differentiating \(F+\ln 3\) returns \(1/x\), confirming the comparison. The absolute values are essential if the comparison is made on the negative interval.
Worked Example: Two Logarithms of Quadratic Expressions
Compare the functions \(F(x)=\ln(x^2+x+1)\) and \(G(x)=\ln((2x+1)^2+3)\). Their logarithm arguments are positive for every real \(x\):
Differentiate \(F\):
For \(G\), the derivative of its inner expression is \(4(2x+1)\). Also, \((2x+1)^2+3=4(x^2+x+1)\). Hence
The relationship between the expressions also identifies the constant difference:
Thus both formulas are antiderivatives of \((2x+1)/(x^2+x+1)\) on \(\mathbb R\), and they differ by \(\ln 4\). The derivative calculation and the logarithm identity provide independent checks.
Equivalent Expressions on Different Intervals
Logarithm identities involving absolute values can combine terms, but their domains must be checked. For nonzero real numbers \(u\) and \(v\), \(\ln|u|+\ln|v|=\ln|uv|\). This identity is valid because \(|u||v|=|uv|\), and both logarithm arguments are positive.
Worked Example: Combining Logarithmic Terms
Consider \(f(x)=1/x+1/(x+1)\), which is defined for \(x\ne0,-1\). On the interval \(I=(0,\infty)\), define
Differentiate \(F\) term by term:
Since \(x(x+1)>0\) on \(I\), differentiating \(G\) gives
Moreover, the product identity gives \(G=F\) on \(I\), so the difference is the constant zero. As a final check, the displayed derivatives of both forms equal the original integrand.
The formulas in this example also work on \((-1,0)\), since neither \(x\) nor \(x+1\) is zero there and the absolute values are defined. The derivative comparison applies separately on that interval. It does not combine the two intervals into one domain: the excluded points \(0\) and \(-1\) divide the domain into three pieces.
A Common Pitfall: Ignoring Domains
A common error is to use \(\ln x\) as an antiderivative of \(1/x\) on a negative interval. For example, on \((-2,-1)\), the expression \(\ln x\) is not real-valued, so the proposed antiderivative is not defined there. The correction is \(\ln|x|\): it is defined throughout that interval and its derivative is \(1/x\).
Another related mistake is to conclude that two functions with the same derivative on a disconnected set must differ by one constant on the whole set. For instance, let the domain be \((-\infty,0)\cup(0,\infty)\), and define \(F(x)=0\) everywhere, while \(G(x)=0\) for \(x<0\) and \(G(x)=1\) for \(x>0\). Both derivatives are zero at every point in the domain, but \(F-G\) is not one constant across the entire domain. It is constant on each interval component, with different values on the two components.
Check Your Understanding
For each comparison, state the relevant interval or domain and explain how differentiation or subtraction verifies the relationship.
- Compare \(x^2-4x\) and \((x-2)^2\). What is their constant difference?
- On \(x\ne0\), compare \(\ln|5x|\) and \(\ln|x|\). What constant separates them?
- Why does equality of derivatives imply a constant difference on an interval?
- Why is \(\ln x\) not an appropriate real antiderivative of \(1/x\) on \((-\infty,0)\)?
- What conclusion is justified if two derivatives agree on a domain with multiple interval components?
