Selecting Substitutions by Inner-Function Fit

Choosing a Substitution by Its Derivative

When an integrand contains a composite expression, a substitution can simplify it by treating the inner expression as a new variable. The key question is not simply whether an expression looks like an inner function. It is whether the other factors in the integrand match its derivative.

For instance, if a power such as \((3x^2+2)^5\) appears, then \(3x^2+2\) is a natural candidate for a new variable. Its derivative is \(6x\). If the integrand also contains \(6x\), the substitution fits directly: the power becomes a power of the new variable, and the remaining factor becomes its differential.

This derivative match is the basis of substitution selection. It tells us when substitution turns the integrand into a simpler expression, and when a tempting choice is missing a necessary factor.

What Counts as a Fit?

Suppose an integrand can be written as \(F(g(x))g'(x)\), where \(g\) is differentiable and \(F\) has an antiderivative \(H\) on the relevant interval. Setting \(u=g(x)\) gives \(du=g'(x)\,dx\), so the integral becomes \(\int F(u)\,du\). The substitution works because it replaces the composite inner expression and its derivative together.

A constant multiple also fits. If the integrand is \(F(g(x))h(x)\), and \(h(x)=c\,g'(x)\) for a nonzero constant \(c\), then the original integrand is \(cF(g(x))g'(x)\). The constant can be carried through the substitution.

The fit need not be an exact visual copy of the derivative. What matters is whether the factors can be rearranged or a constant can be taken out to produce \(g'(x)\,dx\). But if a factor is missing or cannot be obtained by such a rearrangement, the substitution does not directly reduce the integral to a simpler one.

Why the Derivative Match Works

The justification is the chain rule in reverse. Let \(H\) be an antiderivative of \(F\), so \(H'(u)=F(u)\). Differentiating the composite function \(H(g(x))\) gives:

Thus \(H(g(x))\) is an antiderivative of \(F(g(x))g'(x)\). This establishes the substitution rule for indefinite integrals:

If instead the integrand contains \(c\,g'(x)\), the constant factor changes the result accordingly:

In practice, this means we can choose a candidate \(g(x)\), calculate \(g'(x)\), and compare that derivative with the remaining factors. If they match up to a constant, substitution is promising. If they do not, we should not proceed as though they do.

Worked Examples

How to Test a Candidate Efficiently

A useful test has three parts. First, identify a plausible inner function, often the expression inside a power, exponential, or denominator. Second, differentiate it. Third, compare that derivative with the factors outside the composite expression. This keeps the choice tied to the actual structure of the integrand rather than to appearance alone.

The derivative comparison can be especially revealing in common patterns:

These are signals to test, not automatic instructions. A candidate is useful only if its derivative is accounted for by the rest of the integrand. In the rational pattern, for example, a numerator equal to the denominator’s derivative produces the form \(\int \frac{1}{u}\,du\). A different numerator may not.

A Common Mistake: Forgetting the Constant

Consider \(\int \frac{x}{x^2+1}\,dx\). The denominator suggests \(u=x^2+1\), but its derivative is \(2x\), not \(x\). A common error is to write \(du=x\,dx\) and conclude that the integral is \(\ln(x^2+1)+C\). That substitution equation is incorrect: differentiating \(x^2+1\) gives \(2x\).

The correction is to account for the factor of \(2\):

Now the integral is:

Since \(x^2+1>0\), the absolute-value signs are unnecessary in the final expression. Differentiate to check:

The factor of \(\frac{1}{2}\) is essential. Matching a derivative up to a constant does not mean the constant can be ignored; it must be carried through the substitution.

When a Candidate Does Not Fit

Not every composite expression has a useful derivative match. For example, in \(\int e^{x^2}\,dx\), choosing \(u=x^2\) gives \(du=2x\,dx\), but the integrand contains no factor of \(x\). The substitution therefore does not turn the integral directly into \(\int e^u\,du\). This observation does not prove that no method can handle an integral; it only shows that this particular direct substitution lacks its required derivative factor.

Conversely, a match may be hidden in the way the integrand is written. Before rejecting a candidate, check whether a constant can be factored out or the expression can be rearranged without changing it. Once the derivative fit is established, substitution is mechanical; choosing the candidate carefully is the strategic step.