Why Check an Indefinite Integral?
An indefinite integral found by substitution is a proposed antiderivative. The most direct way to test it is to differentiate it and compare the result with the original integrand. This check is especially useful when the expression has several layers: a correct inner function with a missing factor can make an answer look plausible while still making it wrong.
The check reverses the calculation. Substitution uses the chain rule in reverse to construct a candidate; differentiation applies the chain rule forward to test that candidate. The required comparison is exact: the derivative must equal the original integrand at every point in the domain being considered.
The interval and domain matter. A formula may be valid only where each expression in it is defined. In particular, a logarithm such as \(\ln|g(x)|\) requires \(g(x)\ne 0\); a verification applies on an interval on which that condition holds.
A Chain-Rule Test
Suppose a substitution identifies an inner function \(g(x)\) and a corresponding function \(G\) satisfying \(G'(u)=h(u)\). The chain rule gives the derivative of the composed candidate \(G(g(x))\): differentiate the outer function, evaluate at the inner function, and multiply by the derivative of the inner function.
This formula gives a direct check for a substitution-based answer. If the original integrand is \(h(g(x))g'(x)\), then \(G(g(x))\) differentiates back to it. If the integrand instead includes a constant multiple, that factor must also appear in the proposed antiderivative.
The arbitrary constant in an indefinite integral does not affect the check, because the derivative of a constant is zero. It is customary to include \(+C\) when writing the family of antiderivatives, but the derivative test is applied to the nonconstant part.
Worked Example: A Power of a Composite Function
Worked Example: A Power of a Composite Function
Check the proposed result for \(\int 6x(3x^2+1)^4\,dx\).
Take the proposed antiderivative to be \(F(x)=\frac{1}{5}(3x^2+1)^5+C\). The inner function is \(g(x)=3x^2+1\), and its derivative is \(g'(x)=6x\). Differentiate \(F\) using the chain rule:
This is exactly the original integrand, so the proposed antiderivative checks. Therefore, on \(\mathbb{R}\),
As a final comparison, the original integrand has the factor \(6x\), and the differentiated answer has the same factor. The constant \(C\) contributes zero.
Worked Example: A Logarithmic Antiderivative
Worked Example: A Logarithmic Antiderivative
Check the proposed result for \(\int \frac{2x}{x^2+4}\,dx\). Since \(x^2+4>0\) for every real \(x\), the candidate \(\ln(x^2+4)+C\) is defined on all of \(\mathbb{R}\).
Differentiate the proposed antiderivative. The derivative of \(\ln v\) with respect to \(v\) is \(1/v\), so the chain rule gives
The result agrees exactly with the integrand. Hence
The check also shows why the numerator matters: differentiating the logarithm supplies the denominator \(x^2+4\), while differentiating its inner function supplies the numerator \(2x\).
Worked Example: Checking a Logarithm with an Absolute Value
Worked Example: Checking a Logarithm with an Absolute Value
Check the proposed result for \(\int \frac{3x^2+2x}{x^3+x^2+5}\,dx\). Write \(q(x)=x^3+x^2+5\). The proposed antiderivative is \(\ln|q(x)|+C\), which is defined wherever \(q(x)\ne 0\).
On any interval where \(q(x)\ne 0\), the derivative of \(\ln|q(x)|\) is \(q'(x)/q(x)\). Here,
Now differentiate the proposed answer and substitute this derivative:
This recovers the original integrand at every point where the denominator is nonzero. Thus the check confirms the formula on each interval in the domain:
The absolute value allows the logarithm to be defined when \(q(x)\) is negative as well as when it is positive; it does not allow \(q(x)=0\). The verification therefore always retains the nonzero-denominator condition.
Why the Verification Works
The method is justified by the chain rule, rather than by a visual resemblance between the answer and the integrand. The following argument states exactly what must match.
Suppose \(G\) is differentiable and \(g\) is differentiable on an interval \(I\). Consider \(F(x)=G(g(x))\).
By the chain rule, \(F'(x)=G'(g(x))g'(x)\) for every \(x\in I\).
If \(f(x)=G'(g(x))g'(x)\) throughout \(I\), then \(F'(x)=f(x)\) throughout \(I\). By the definition of antiderivative, \(F\) is an antiderivative of \(f\) on \(I\).
This is a complete verification: once the derivative equality holds throughout the stated interval, the definition itself establishes that the candidate is an antiderivative. If a constant \(C\) is added to \(F\), then \((F+C)'=F'+0=f\), so every such constant gives another antiderivative.
On an interval, any two antiderivatives of the same function differ by a constant. To see why, let \(F\) and \(H\) both be antiderivatives of \(f\) on \(I\). Then \((F-H)'=F'-H'=f-f=0\). For any \(a,b\in I\) with \(a<b\), the mean value theorem applied to \(F-H\) on \([a,b]\) gives some \(c\in(a,b)\) such that \((F-H)(b)-(F-H)(a)=(F-H)'(c)(b-a)=0\). Therefore \((F-H)(b)=(F-H)(a)\), so \(F-H\) is constant on \(I\). This explains why the indefinite integral is written with \(+C\).
A Common Mistake: Losing the Inner Derivative
A frequent error is to differentiate only the outside of a composition. For example, one might incorrectly write
This omits the derivative of the inner function \(3x^2+1\). The chain rule requires that factor:
Consequently, if the proposed antiderivative for \(6x(3x^2+1)^4\) were \((3x^2+1)^5\), its derivative would be \(30x(3x^2+1)^4\), not \(6x(3x^2+1)^4\). The candidate is too large by a factor of \(5\). Including the multiplier \(\frac{1}{5}\) corrects it:
The comparison with the original integrand exposes the error immediately. A substitution may reveal the right outer form, but the derivative test confirms whether the coefficient is correct.
A Reliable Checking Routine
When checking an answer obtained by substitution, carry out these steps in order:
- Write the proposed antiderivative without its \(+C\), which has derivative zero.
- Differentiate the outer function and multiply by the derivative of every inner function required by the chain rule.
- Keep any constant multiplier in the proposed answer; do not discard it during differentiation.
- Simplify the derivative and compare it with the original integrand exactly.
- State any domain restriction, such as a denominator being nonzero or a logarithm's argument being nonzero.
This routine is useful because it separates two questions: whether the candidate has the correct form, and whether its derivative truly equals the given function. Only the second question establishes that it is an antiderivative.
Check Your Understanding
For each question, use differentiation to decide whether the proposed expression is an antiderivative on the stated domain.
- Differentiate \(\frac{1}{3}(2x+1)^3\). What integrand does it reproduce?
- Why does \(\ln|x^2-9|\) require a domain restriction when used as an antiderivative?
- A proposed answer to \(\int 4x e^{2x^2}\,dx\) is \(e^{2x^2}+C\). Differentiate it and identify any missing factor.
- Explain why adding \(C\) to a verified antiderivative does not change the derivative test.
- If two functions are antiderivatives of the same function on an interval, what does their difference equal?
