Why Basic Integration Errors Are Easy to Miss
Many incorrect antiderivatives look plausible. A power may appear to change in the right direction while its coefficient is wrong; a minus sign may disappear; or a proposed answer may differentiate correctly but omit most of the antiderivatives. The reliable check is to differentiate the proposed answer and compare it with the original integrand.
This tutorial focuses on five recurring sources of error: changing powers and coefficients, treating the reciprocal as an ordinary power, integrating an exponential in a denominator, remembering trigonometric signs, and including the constant of integration.
The Power Rule—and Its Exception
For an integer \(n\ne -1\), the power rule for antiderivatives is
Here \(C\) is an arbitrary constant. If \(n\) is negative, the expression is defined only where its denominator is nonzero, so the rule is applied on intervals that avoid \(x=0\). For nonnegative integer powers, the formula is defined for all real \(x\).
The exponent increases by one, and the new power is divided by that increased exponent. Both steps matter. For example, the antiderivative of \(x^4\) is \(x^5/5\), not \(x^5\) or \(5x^5\).
The exponent \(n=-1\) is the exception: adding one would give an exponent of zero, and the denominator \(n+1\) would be zero. Instead, on any interval that does not contain zero,
The absolute value is important because \(\ln|x|\) is defined for both positive and negative \(x\), while \(1/x\) is defined on both sides of zero. The interval restriction still matters: an antiderivative on a positive interval and one on a negative interval may have different constants.
Worked Example: Separating Ordinary Powers from the Reciprocal
Find an antiderivative of \(3x^2-4x^{-2}+2/x\) on an interval that does not contain zero.
Integrate each term separately. For \(3x^2\), increase the exponent from \(2\) to \(3\), then divide by \(3\):
For \(-4x^{-2}\), increase the exponent from \(-2\) to \(-1\), then divide by \(-1\):
The final term is \(2x^{-1}\), the exceptional reciprocal power, so it uses the logarithm rule:
Thus an antiderivative is \(F(x)=x^3+4/x+2\ln|x|+C\). Check by differentiating each term:
The derivative equals the original integrand, so the antiderivative is correct on each interval where \(x\ne0\).
Exponential Functions in Denominators
An exponential in a denominator can be rewritten using a negative exponent: \(1/e^{kx}=e^{-kx}\). This makes the sign of the exponent visible before you integrate. For \(k\ne0\), differentiating \(e^{-kx}\) gives \(-k e^{-kx}\), so the antiderivative must include the factor \(-1/k\):
This formula is a common place to lose a minus sign or a coefficient. The denominator form does not make the integral positive by itself: the negative exponent has derivative \(-k e^{-kx}\).
Worked Example: Integrating an Exponential Denominator
Find \(\int 5/e^{2x}\,dx\). First rewrite the integrand so that the exponent is explicit:
Since the exponent is \(-2x\), an antiderivative is obtained by dividing by \(-2\):
Check by differentiating:
The two negative factors in the differentiation cancel, recovering the positive integrand.
Trigonometric Signs Must Be Checked
Trigonometric antiderivatives are especially vulnerable to sign errors because differentiation changes the sign of cosine but not sine in the same way. The basic identities to check are:
| Integrand | Antiderivative | Sign check |
|---|---|---|
| \(\sin x\) | \(-\cos x+C\) | \(\frac{d}{dx}(-\cos x)=\sin x\) |
| \(\cos x\) | \(\sin x+C\) | \(\frac{d}{dx}(\sin x)=\cos x\) |
| \(\sec^2 x\) | \(\tan x+C\) | \(\frac{d}{dx}(\tan x)=\sec^2 x\) |
| \(\csc^2 x\) | \(-\cot x+C\) | \(\frac{d}{dx}(-\cot x)=\csc^2 x\) |
These formulas are used on intervals where the functions involved are defined. For example, \(\tan x\) is not defined where \(\cos x=0\), and \(\cot x\) is not defined where \(\sin x=0\). The sign can be checked locally by differentiating; memorizing a sign without checking it is less reliable.
Worked Example: Checking Signs in a Trigonometric Integral
Find an antiderivative of \(4\sin x-3\cos x\).
Use \(\int\sin x\,dx=-\cos x+C\) and \(\int\cos x\,dx=\sin x+C\), retaining the coefficient of each term:
Differentiate the result term by term:
This matches the integrand. In particular, the minus sign before \(4\cos x\) is necessary: differentiating \(4\cos x\) would give \(-4\sin x\), the wrong sign.
Why the Constant Belongs in the Answer
If \(F'(x)=f(x)\), then for any constant \(C\),
Therefore, an indefinite integral must describe a family, not just one function. The constant is not an optional decoration: leaving it out gives one antiderivative but fails to state all antiderivatives on the interval.
There is also a reason there are no other differences between antiderivatives on the same interval. Suppose \(F'=f\) and \(G'=f\) there. Then \(H=F-G\) satisfies \(H'=0\). For any two points \(a<b\) in the interval, the Mean Value Theorem gives a point \(c\in(a,b)\) such that
Thus \(H(a)=H(b)\) for every such pair, so \(H\) is constant on the interval. Hence \(G=F+C\). This proves that all antiderivatives on an interval differ only by a constant.
Worked Example: Correcting an Omitted Constant
Consider \(\int(2x+1)\,dx\). Integrating the two terms gives \(x^2+x\), but this is only one antiderivative. The complete family is
Check the result:
For example, \(x^2+x\) and \(x^2+x+7\) both differentiate to \(2x+1\). Omitting \(C\) reports only the first of these possibilities, rather than the entire family.
A Reliable Error-Checking Routine
When checking a basic antiderivative, do not rely only on whether the expression looks familiar. Differentiate it and compare the result with the original integrand. The following routine targets the errors discussed here:
For a power \(x^n\), confirm that the exponent increases by one and the result is divided by the new exponent. If \(n=-1\), use \(\ln|x|\) instead.
Express \(1/e^{kx}\) as \(e^{-kx}\), then check that the coefficient accounts for differentiating the exponent \(-kx\).
Differentiate the proposed trigonometric antiderivative; a sign error will remain visible in the resulting term.
For an indefinite integral on an interval, write \(+C\) to represent the full family of antiderivatives.
Confirm that every term and coefficient reproduces the original integrand.
Common Mistakes and Their Corrections
A frequent power-rule error is to apply the ordinary formula to \(1/x\): someone may write \(\int x^{-1}\,dx=x^0/0\). This is undefined because the proposed denominator is zero. The correction is to recognize the exceptional exponent and use \(\ln|x|+C\) on an interval not containing zero.
Another common error is to write \(\int e^{-2x}\,dx=e^{-2x}+C\). Differentiating that proposed answer gives \(-2e^{-2x}\), not \(e^{-2x}\). Dividing by \(-2\) corrects both the sign and coefficient: \(-\frac12e^{-2x}+C\).
For trigonometric functions, a wrong sign can be exposed just as directly. The incorrect claim \(\int\sin x\,dx=\cos x+C\) differentiates to \(-\sin x\). The correction is \(-\cos x+C\), whose derivative is \(\sin x\). Finally, an answer without \(C\) may differentiate correctly, but it does not describe every antiderivative.
Check Your Understanding
For each question, pay attention to the exponent, sign, domain, and constant as appropriate.
- What is \(\int x^{-4}\,dx\), and how can differentiation verify your answer?
- Why does the ordinary power rule fail for \(\int x^{-1}\,dx\), and what formula replaces it?
- Find an antiderivative of \(3/e^{3x}\) and check it by differentiation.
- What is the sign in an antiderivative of \(\sin x\)? Verify the sign by differentiating.
- Why does \(x^2\) give only one antiderivative of \(2x\), rather than the complete family?
