Introduction
A mixed expression can look like one calculation, but its terms may come from several different derivative rules. The key is to recognize each term separately: ask what familiar function would differentiate to produce it, and then assemble those functions into an antiderivative of the whole expression.
This method depends on matching accurately. A coefficient may need to be divided out, a negative sign may change the antiderivative, and some formulas are valid only on specified domains. Differentiation provides a direct check that the pieces have been matched correctly.
What It Means to Match a Derivative Pattern
A function \(F\) is an antiderivative of \(f\) on an interval \(I\) if \(F\) is differentiable there and \(F'(x)=f(x)\) for every \(x\) in \(I\). In a mixed expression, we look at each term of \(f\) and compare it with a familiar derivative.
The basic rules most useful for this task are summarized below. Each rule gives a function whose derivative is the expression in the left column. In the power rule, \(n\neq -1\); the reciprocal case is listed separately.
| Expression to match | Corresponding antiderivative form | Conditions |
|---|---|---|
| \(x^n\) | \(\frac{x^{n+1}}{n+1}\) | \(n\neq -1\) |
| \(\frac{1}{x}\) | \(\ln|x|\) | \(x\neq 0\) |
| \(e^{kx}\) | \(\frac{1}{k}e^{kx}\) | \(k\neq 0\) |
| \(a^x\) | \(\frac{a^x}{\ln a}\) | \(a>0,\ a\neq 1\) |
| \(\sin x\) | \(-\cos x\) | All real \(x\) |
| \(\cos x\) | \(\sin x\) | All real \(x\) |
| \(\sec^2 x\) | \(\tan x\) | Where defined |
| \(\csc^2 x\) | \(-\cot x\) | Where defined |
| \(\sec x\tan x\) | \(\sec x\) | Where defined |
These are matching rules, not instructions to ignore constants. For instance, if a term is \(7\cos x\), its matching antiderivative is \(7\sin x\). If the term is \(7e^{3x}\), its matching antiderivative is \(\frac{7}{3}e^{3x}\), because differentiating \(e^{3x}\) produces an extra factor of \(3\).
How to Match Terms Systematically
For each term, first identify its main form: a power, reciprocal, exponential, or trigonometric expression. Then check the coefficient and sign against the derivative rule. Finally, write a candidate antiderivative for that term. Once all terms have been handled, add the candidates and include a constant of integration.
Use the plus and minus signs to identify the individual pieces of the expression.
Compare each piece with a basic derivative form, including any coefficient inside an exponential.
Adjust its coefficient or sign so that differentiating it gives the original term exactly.
Add the candidates, include \(C\), and differentiate the sum term by term.
The constant \(C\) is included because adding a constant does not change a derivative. On a connected interval, any two antiderivatives of the same function differ by a constant. If the domain has more than one connected component, constants may be chosen independently on those components.
Worked Example: Powers, an Exponential, and a Reciprocal
Worked Example: Powers, an Exponential, and a Reciprocal
Find an antiderivative of \(6x^5-4e^{-2x}+3\cos x-\frac{2}{x}\) on the interval \(x>0\).
Match the first term with the power rule. Since \(\frac{d}{dx}(x^6)=6x^5\), the first term has antiderivative \(x^6\).
For the exponential, \(\frac{d}{dx}(e^{-2x})=-2e^{-2x}\). Therefore, \(2e^{-2x}\) differentiates to \(-4e^{-2x}\).
Since \(\frac{d}{dx}(\sin x)=\cos x\), the third term has antiderivative \(3\sin x\). On \(x>0\), \(\ln|x|=\ln x\), and \(\frac{d}{dx}(\ln x)=1/x\); hence the final term has antiderivative \(-2\ln x\).
Check by differentiating each term:
This is the original expression, so the proposed antiderivative is verified on \(x>0\).
Why Termwise Matching Works
The method is justified by the linearity of differentiation. If \(F_1'(x)=f_1(x)\) and \(F_2'(x)=f_2(x)\), then the sum rule for derivatives gives \((F_1+F_2)'=f_1+f_2\). The same principle applies to any finite number of terms and to constant multiples.
This justification also explains why verification is effective: differentiation reverses each match, and the sum rule restores the original combination of terms. It does not mean that every expression can be integrated by these basic patterns. If a term does not match a rule being used, this method alone does not supply its antiderivative.
Worked Example: Exponential and Trigonometric Patterns
Worked Example: Exponential and Trigonometric Patterns
Find an antiderivative of \(5\cdot 2^x+4\sin x-3\csc^2 x+7\sec x\tan x\) on an interval where the trigonometric terms are defined.
Because \(\frac{d}{dx}(2^x)=2^x\ln 2\), the matching antiderivative of \(5\cdot 2^x\) is \(\frac{5\cdot 2^x}{\ln 2}\). The term \(4\sin x\) matches \(-4\cos x\), since the derivative of \(\cos x\) is \(-\sin x\). As \(\frac{d}{dx}(\cot x)=-\csc^2 x\), the third term matches \(3\cot x\). Finally, \(\frac{d}{dx}(\sec x)=\sec x\tan x\), so the last term matches \(7\sec x\).
Differentiate to check every term:
The check reproduces the given expression. The interval must avoid points where \(\sin x=0\) or \(\cos x=0\), so that the trigonometric terms and the proposed antiderivative are defined throughout it.
Recognizing Logarithmic Trigonometric Forms
Some basic trigonometric derivative patterns lead to logarithms. Where \(\cos x\neq 0\), \(\frac{d}{dx}\ln|\cos x|=-\tan x\), so an antiderivative of \(\tan x\) is \(-\ln|\cos x|\). Where \(\sin x\neq 0\), \(\frac{d}{dx}\ln|\sin x|=\cot x\), so an antiderivative of \(\cot x\) is \(\ln|\sin x|\). The absolute values allow the logarithm to be defined on intervals where the sine or cosine is negative as well as positive.
Worked Example: Tangent, Cotangent, and a Power
Find an antiderivative of \(2\tan x-3\cot x+4x^2\) on an interval where both \(\sin x\) and \(\cos x\) are nonzero.
Match \(2\tan x\) with \(-2\ln|\cos x|\), and match \(-3\cot x\) with \(-3\ln|\sin x|\). By the power rule, \(4x^2\) matches \(\frac{4}{3}x^3\). Thus a candidate is
Check by differentiation. The derivative of \(-2\ln|\cos x|\) is \(2\tan x\), and the derivative of \(-3\ln|\sin x|\) is \(-3\cot x\). Also, \(\frac{d}{dx}(\frac{4}{3}x^3)=4x^2\). Therefore,
This is exactly the given expression, which verifies the result on any interval satisfying the stated domain condition.
Common Mistake: Forgetting the Factor from the Exponent
A frequent error is to treat \(e^{kx}\) as if its derivative were just \(e^{kx}\). For example, the calculation \(\int 6e^{-2x}\,dx=6e^{-2x}+C\) is incorrect: differentiating the proposed answer gives \(-12e^{-2x}\), not \(6e^{-2x}\).
The derivative of \(e^{-2x}\) is \(-2e^{-2x}\). To obtain \(6e^{-2x}\), the coefficient in the antiderivative must be \(-3\), since
When a match seems plausible, differentiating the candidate is the safest way to catch a missing factor or sign before combining it with the other terms.
Using the Method Reliably
Matching is most reliable when each term is checked independently. Do not let the appearance of a whole expression distract from a term that has a different pattern. In particular, \(x^{-1}\) is the exceptional power that does not use the ordinary power rule; its basic antiderivative is \(\ln|x|\). Exponentials with \(kx\) in the exponent require division by \(k\), and trigonometric signs should be checked against the derivative of the proposed function.
After writing a candidate sum, differentiate it before treating the calculation as complete. The check must reproduce every term, including its coefficient and sign. If it does not, compare the mismatched term with the derivative rule that was used and correct that match.
Check Your Understanding
For each question, use derivative patterns and check the proposed result by differentiating where appropriate.
- What basic antiderivative pattern matches \(8x^7\)?
- What coefficient must multiply \(e^{4x}\) to give an antiderivative of \(12e^{4x}\)?
- Give an antiderivative of \(-5\sin x\), and state how differentiation checks its sign.
- Why is \(\ln|x|\), rather than the ordinary power rule, used for \(1/x\)?
- Find and verify an antiderivative of \(3\sec x\tan x+2\cos x\) on an interval where the terms are defined.
