Why Checking the Answer Requires Checking Its Domain
A rationalizing substitution can turn an algebraic integrand into a rational one, making its antiderivative easier to find. But the resulting expression is only an answer to the original problem if it is valid on the interval under consideration and differentiates back to the original integrand there.
The crucial point is that substitutions involving radicals and inverse formulas come with branch restrictions. For example, the principal square root \(\sqrt{x^2-1}\) is nonnegative, while an inverse formula for \(x\) may hold only when its parameter lies in a particular range. Algebra can produce a plausible expression without preserving those restrictions.
There is a direct check: write the proposed antiderivative as a function of the original variable and differentiate it, using the chain rule where needed. Alongside that calculation, check where the original integrand exists and where each expression in the answer is defined.
A Reliable Verification Method
Suppose a substitution gives \(t=t(x)\), with inverse \(x=x(t)\), and the transformed integral has antiderivative \(G(t)\). The candidate answer in the original variable is \(F(x)=G(t(x))\). On an interval where these functions are defined and differentiable, the chain rule gives \(F'(x)=G'(t(x))t'(x)\). The substitution calculation should make this product equal to the original integrand.
This is why differentiating after back-substitution is an effective check: it tests the final formula in the variable and branch of the original problem, rather than relying only on the transformed integral.
Require the radicand to be nonnegative and exclude points where the integrand is undefined, such as zeros of a denominator or points where a reciprocal radical is singular.
State the range of the substitution parameter and the conditions under which the inverse formula and radical formula hold.
Write the candidate antiderivative explicitly in terms of the original variable, preserving absolute values where required.
Use the chain rule and simplify to the original integrand on the stated interval.
Worked Example: A Bounded Radical
Worked Example: Checking an Antiderivative for \(\sqrt{4-x^2}\)
Consider \(\int \frac{dx}{\sqrt{4-x^2}}\). The integrand is defined when \(4-x^2>0\), so its domain is the open interval \((-2,2)\). Use \(t=\frac{x}{2+\sqrt{4-x^2}}\). On this interval, \(t\) lies in \((-1,1)\), and the inverse and radical formulas are \(x=\frac{4t}{1+t^2}\) and \(\sqrt{4-x^2}=\frac{2(1-t^2)}{1+t^2}\).
Differentiate the inverse formula and divide by the radical:
Thus a transformed antiderivative is \(2\arctan t\), giving the candidate \(F(x)=2\arctan\!\left(\frac{x}{2+\sqrt{4-x^2}}\right)\). To check it, put \(s=\sqrt{4-x^2}\). Since \(s'=-x/s\), differentiating \(t=x/(2+s)\) gives
Also, using \(x^2=4-s^2\),
Now differentiate the candidate:
The derivative matches the integrand throughout \((-2,2)\), so \(F\) is an antiderivative on that interval. The endpoints are excluded because the original integrand is undefined there.
Worked Example: A Positive Euler Parameter
Worked Example: Checking a Logarithmic Answer on the Real Line
Consider \(\int \frac{dx}{\sqrt{x^2+4}}\). The radicand is positive for every real \(x\), so the integrand is defined on \(\mathbb{R}\). Let \(t=\frac{x+\sqrt{x^2+4}}{2}\). Since \(\sqrt{x^2+4}>|x|\), the numerator is positive and \(t>0\) for every real \(x\). The inverse relations on this branch are \(x=t-\frac1t\) and \(\sqrt{x^2+4}=t+\frac1t\).
Differentiate the inverse relation and simplify the transformed integrand:
Because \(t>0\), the transformed antiderivative is \(\ln t\). Back-substitution gives \(F(x)=\ln\!\left(\frac{x+\sqrt{x^2+4}}{2}\right)\). To verify it directly, differentiate \(t\):
Therefore,
The logarithm's argument is positive for every real \(x\), exactly as required for this formula. The answer is verified on the whole original domain, not merely for some values of \(x\).
Worked Example: Separate Branches Outside a Radical's Zeros
Worked Example: Checking \(\ln|x+\sqrt{x^2-1}|\)
Consider \(\int \frac{dx}{\sqrt{x^2-1}}\). The radical is real when \(x^2-1\geq 0\), but the reciprocal is undefined at \(x=\pm1\). The integrand's domain is therefore \((-\infty,-1)\cup(1,\infty)\), two separate intervals.
Set \(t=x+\sqrt{x^2-1}\). For \(x>1\), \(t>1\). For \(x<-1\), \(t\) is negative and lies in \((-1,0)\). On either of these parameter ranges, the inverse and radical relations are \(x=\frac12(t+1/t)\) and \(\sqrt{x^2-1}=\frac12(t-1/t)\). Hence
Since \(t\) is nonzero on both branches, an antiderivative with respect to \(t\) is \(\ln|t|\). Thus the back-substituted candidate is \(F(x)=\ln|x+\sqrt{x^2-1}|\). Let \(s=\sqrt{x^2-1}\). Then \(s'=x/s\), so
The derivative of \(\ln|u(x)|\), where \(u(x)\ne0\), is \(u'(x)/u(x)\). Applying that rule here gives
This calculation holds on both intervals. The absolute value is necessary for a real-valued logarithm on the negative branch, where \(x+\sqrt{x^2-1}<0\). As with any antiderivative on a disconnected domain, an independent constant of integration may be chosen on each interval.
Common Pitfall: Extending One Branch Across the Whole Domain
A common error is to use a branch condition derived for one part of the domain everywhere. In the preceding example, the parameter \(t=x+\sqrt{x^2-1}\) satisfies \(t>1\) when \(x>1\), but that condition does not hold when \(x<-1\). For instance, at \(x=-2\), \(t=-2+\sqrt{3}\), which is negative and lies between \(-1\) and \(0\).
Consequently, the expression \(\ln(x+\sqrt{x^2-1})\) is not even real-valued at \(x=-2\). Replacing it by \(\ln|x+\sqrt{x^2-1}|\) gives a real-valued candidate on both intervals, and the derivative check confirms that it works there. The absolute value does not remove the need to check the branch: it makes the logarithm defined once the nonzero argument's sign is understood.
Why the Derivative Check Is Decisive
The substitution method relies on changing variables consistently. If \(G'(t)\) is the transformed integrand and \(t=t(x)\) is valid on an interval, then the chain rule says \(\frac{d}{dx}G(t(x))=G'(t(x))t'(x)\). The transformed integrand was obtained by expressing the original integrand together with \(dx\) in terms of \(t\). Reversing that calculation must recover the original integrand, provided the same branch and interval are used.
That proviso matters. A substitution may be one-to-one only on part of the original domain, its inverse may use a restricted range, or the back-substituted expression may have an excluded point. If any such restriction is ignored, the algebraic formula might be correct on one interval and invalid on another. A derivative check on the claimed interval exposes that mismatch.
Also distinguish an integrand's domain from points at which an antiderivative formula happens to simplify. For a reciprocal radical, a zero of the radicand is still excluded from the integrand even if a transformed or simplified expression has a finite-looking limit there. The domain comes from the original integrand, not from a later algebraic appearance.
Check Your Understanding
Use the domain and branch checks as well as differentiation when answering.
- What two conditions make \(F\) an antiderivative of \(f\) on an open interval?
- For \(\int \frac{dx}{\sqrt{4-x^2}}\), why must the endpoints \(x=\pm2\) be excluded?
- For \(t=\frac{x+\sqrt{x^2+4}}{2}\), why is \(\ln t\) a real-valued expression for every real \(x\)?
- For \(t=x+\sqrt{x^2-1}\), what are the parameter ranges on \(x>1\) and \(x<-1\)?
- Why may an antiderivative on a domain with two separate intervals have different constants on those intervals?
