Why the Choice of Inner Function Matters
A substitution can turn a complicated-looking integral into a familiar one, but its success depends on choosing the right expression to represent with a new variable. The useful expression is often repeated, or it appears inside a power, root, or denominator. The choice must also make sense of the rest of the integrand: after the substitution, every factor should be rewritten in terms of the new variable and its differential.
For example, in an expression involving \(x^2+3\), it is natural to consider naming \(x^2+3\) as \(u\). But recognizing that expression is only the beginning. We must also account for the other factors involving \(x\), and see whether the integral becomes one we know how to evaluate.
What Makes a Substitution Useful?
Let \(u=g(x)\), where \(g\) is a differentiable expression. The substitution links changes in \(x\) and \(u\) through \(du=g'(x)\,dx\). A useful choice makes it possible to express the integrand, including its differential, in terms of \(u\) and \(du\), with no unresolved \(x\)-factors left over. Ideally, the resulting integral is one whose antiderivative is already familiar.
There may be several expressions that look like possible candidates. A good choice often has one or more of these features:
- It is the base of a power, or the expression inside a root or denominator.
- It occurs repeatedly, so replacing it removes repeated complexity.
- The remaining \(x\)-factors can be combined with \(dx\) and rewritten using \(du\).
- Any leftover powers of \(x\) can be expressed in terms of \(u\) using the substitution equation.
These are clues, not a rule that every problem must follow the same pattern. The practical test is whether the entire integral becomes simpler after the substitution. A choice that replaces one complicated expression but leaves several unrelated \(x\)-factors may not help.
Why Substitution Works
The reasoning comes from the chain rule. Suppose an integrand can be written as \(G(g(x))g'(x)\), and suppose \(F\) is an antiderivative of \(G\), so \(F'(u)=G(u)\). Then the chain rule gives the following derivative:
Therefore \(F(g(x))\) is an antiderivative of the original integrand. Substitution packages the repeated inner expression as \(u=g(x)\), so the integral is evaluated as an integral in \(u\), then expressed back in \(x\) for an indefinite integral.
This also explains why choosing an inner function without considering the remaining factors is incomplete. The expression \(g(x)\) identifies what may become \(u\), but the rest of the integrand must fit the rewritten integral too.
A Reliable Selection Process
Use the following process when deciding what to substitute. It is a way to test a choice, not a requirement to use the first expression that catches your eye.
Look for a repeated expression or one contained inside a power, root, or denominator.
Write \(u=g(x)\), then find \(du=g'(x)\,dx\).
Use the equation for \(du\), and, when helpful, solve \(u=g(x)\) for a remaining expression involving \(x\).
If the full integral is now simpler and uses only \(u\) and \(du\), proceed. If important \(x\)-factors remain and cannot be conveniently rewritten, reconsider the choice.
Worked Examples
Worked Example: A Power of a Repeated Quadratic
Choose a substitution and evaluate \(\displaystyle \int 2x(x^2+3)^4\,dx\).
The expression \(x^2+3\) is repeated through its fourth power, so let
Differentiating gives \(du=2x\,dx\). The factor \(2x\,dx\) in the integral is exactly \(du\), and the fourth power becomes \(u^4\). Thus
Since \(u=x^2+3\), the antiderivative in the original variable is
Check by differentiating:
This returns the original integrand, so the antiderivative is correct.
Worked Example: The Denominator and the Remaining Factor
Evaluate \(\displaystyle \int \frac{2x+1}{x^2+x+4}\,dx\).
The quadratic \(x^2+x+4\) is the denominator, making it a natural candidate for \(u\). Its derivative is \(2x+1\), which is precisely the numerator:
Both the denominator and the numerator together with \(dx\) are now expressed using \(u\) and \(du\):
Substitute back to obtain \(\ln|x^2+x+4|+C\). In this case the absolute value is unnecessary for real \(x\): completing the square gives
Thus the answer may be written \(\ln(x^2+x+4)+C\). Check by differentiating:
The derivative matches the original integrand.
Worked Example: Rewriting a Leftover Factor
Evaluate \(\displaystyle \int x\sqrt{9-x^2}\,dx\).
The expression under the square root is \(9-x^2\), so choose
Then \(du=-2x\,dx\), so \(x\,dx=-\frac12\,du\). The remaining factor \(x\) is not left behind: together with \(dx\), it is rewritten using \(du\). Therefore
Replacing \(u\) gives \(-\frac13(9-x^2)^{3/2}+C\). For the original integrand to be real, it is enough to consider \(x\) in the interval \([-3,3]\). Check by differentiating within the interior of that interval:
The result differentiates to the original integrand.
When a Candidate Does Not Help
A substitution is not useful merely because it names an expression in the integral. For instance, in \(\int 2x(x^2+3)^4\,dx\), one could set \(u=x\). Then \(du=dx\), and the integral becomes \(\int 2u(u^2+3)^4\,du\). This is still an integral with essentially the same structure; the substitution has not simplified the repeated quadratic. Choosing \(u=x^2+3\), by contrast, turns the entire expression into \(\int u^4\,du\).
When a first choice leaves \(x\)-factors behind, ask whether they can be rewritten from \(u=g(x)\). For example, if \(u=x^2+3\), then \(x^2=u-3\). This can help when an integral contains an even power of \(x\). If a lone \(x\) remains and the differential does not account for it, rewriting \(x^2\) does not remove that lone factor. The proposed substitution may therefore fail to simplify the integral.
Common Pitfall: Losing a Factor
Consider \(\int \frac{x}{x^2+4}\,dx\), and set \(u=x^2+4\). Then \(du=2x\,dx\), so \(x\,dx=\frac12du\). A common incorrect step is to replace the integral with \(\int \frac1u\,du\), forgetting the factor \(\frac12\).
Differentiation exposes the error:
which is twice the original integrand. The correct rewriting retains the factor:
Here \(x^2+4>0\), so the absolute value can be omitted. Differentiating the corrected answer gives \(\frac12\cdot\frac{2x}{x^2+4}=\frac{x}{x^2+4}\), as required. The lesson is not simply to spot the denominator: preserve the exact relationship between the old differential and \(du\).
What to Check Before Integrating
Before carrying out an antiderivative, pause after choosing \(u\) and inspect the entire rewritten integral. In particular, check that:
- the expression named \(u\) is the one you intended to replace;
- the differential \(du\) has been calculated correctly;
- all remaining factors, including constants, have been rewritten;
- the new integral is actually simpler than the original.
A useful substitution is a choice that organizes the whole integrand, not just one part of it. Once the choice has been made and the factors have been rewritten consistently, familiar antiderivative rules can finish the calculation. Differentiating the final answer checks that the substitution and the integration have both been handled correctly.
Check Your Understanding
For each question, focus on whether the proposed substitution accounts for the whole integrand.
- For \(\int 6x(x^2+5)^2\,dx\), what inner expression would you choose, and what does the integral become in the new variable?
- For \(\int \frac{3x^2}{x^3+1}\,dx\), identify a natural candidate for \(u\). Which part of the integrand becomes \(du\)?
- For \(\int x\sqrt{16-x^2}\,dx\), what substitution turns the expression under the root into \(u\), and how do you rewrite \(x\,dx\)?
- Why is it not enough to replace a repeated expression with \(u\) while leaving an unaccounted factor of \(x\) in the integral?
- In \(\int \frac{x}{x^2+4}\,dx\), what constant factor appears when \(x\,dx\) is rewritten in terms of \(du\)?
