Why the Differential Must Change
A substitution changes more than the expression inside an integrand. When \(u=g(x)\), the differential changes with it: \(du=g'(x)\,dx\). This relationship is where the necessary constants and factors come from.
A missing factor changes the value of an antiderivative; an extra factor does too. The safest approach is to write the differential relationship first, then identify exactly which part of the original integrand it replaces. This tutorial concentrates on that bookkeeping for indefinite integrals.
Read the Differential Relationship Literally
Suppose \(u=g(x)\) and \(g\) is differentiable. Differentiating both sides with respect to \(x\) gives
In substitution notation, this is written as \(du=g'(x)\,dx\). It records how the change in \(u\) relates to the change in \(x\). The notation is especially useful when the integrand contains \(g'(x)\) or a constant multiple of it.
For instance, if \(u=x^2+5\), then \(du=2x\,dx\). This means \(2x\,dx\) can be replaced by \(du\). If the integral contains only \(x\,dx\), then \(x\,dx=\frac12du\). The factor \(\frac12\) is required because the original differential factor is half of \(du\).
Sometimes the substitution is linear, such as \(u=5x-3\). Then \(du=5\,dx\), so \(dx=\frac15du\). In that case, the factor appears when replacing \(dx\) itself. These are two ways of expressing the same rule: transform the complete differential factor present in the integral.
Matching Factors Without Losing the Constant
A useful pattern is an integrand of the form \(H(g(x))g'(x)\). The expression \(g'(x)\,dx\) is exactly \(du\), so the integral becomes one in \(u\). If the integrand instead contains \(c\,g'(x)\), retain the constant \(c\) after the substitution.
The constant may also appear as a fraction. If a derivative produces \(6x\), but the integrand contains \(2x\,dx\), then \(2x\,dx=\frac13du\). Conversely, if the integral contains \(12x\,dx\), then \(12x\,dx=2du\). The factor is determined by comparing the actual differential factor with \(g'(x)\,dx\), not by guessing.
A practical sequence is:
Set \(u=g(x)\), and compute \(g'(x)\).
Record \(du=g'(x)\,dx\) before replacing any factors.
Express the original differential factor as a multiple of \(du\), if possible.
After integrating in \(u\), substitute \(g(x)\) back and check by differentiation.
Why the Substitution Rule Works
The factor \(g'(x)\) is not an optional adjustment. It is the factor produced when a function of \(g(x)\) is differentiated. The chain rule therefore gives a direct justification for the substitution rule.
Therefore \(F(g(x))\) is an antiderivative of \(H(g(x))g'(x)\). If a constant \(c\) multiplies that integrand, then the chain rule gives
This proves the substitution formula for indefinite integrals: whenever \(F'=H\), the expression \(F(g(x))+C\) is an antiderivative of \(H(g(x))g'(x)\). It also explains why the factors must match. The derivative of the proposed answer will produce \(g'(x)\); if the original integrand has only a fraction or a multiple of that derivative, the same fraction or multiple must remain in the antiderivative.
Worked Examples
Worked Example: The Differential Matches Exactly
Evaluate \(\int 6x(3x^2+2)^4\,dx\).
Choose the inner expression \(u=3x^2+2\). Its derivative is \(\frac{du}{dx}=6x\), so \(du=6x\,dx\). This matches the factor \(6x\,dx\) in the integral exactly.
Check by differentiating the result:
The derivative equals the original integrand, so the antiderivative is correct.
Worked Example: The Differential Is Half of \(du\)
Evaluate \(\int \frac{x}{x^2+9}\,dx\).
Set \(u=x^2+9\). Then \(\frac{du}{dx}=2x\), so \(du=2x\,dx\), or \(x\,dx=\frac12du\). The numerator and differential together contribute only half of \(du\).
Since \(x^2+9>0\) for every real \(x\), this can also be written as \(\frac12\ln(x^2+9)+C\). Check the result:
The factor \(\frac12\) is necessary: it cancels the \(2\) produced by differentiating \(x^2+9\).
Worked Example: Transforming \(dx\) for a Linear Substitution
Evaluate \(\int e^{4x-1}\,dx\).
Let \(u=4x-1\). Then \(\frac{du}{dx}=4\), so \(du=4\,dx\), and therefore \(dx=\frac14du\). There is no separate factor of \(4\) in the original integrand to absorb this constant.
Differentiate to check:
The factor \(\frac14\) in the antiderivative is exactly what cancels the \(4\) from the chain rule.
Worked Example: A Constant Multiple Greater Than One
Evaluate \(\int 4x(1+x^2)^3\,dx\).
Choose \(u=1+x^2\). Then \(\frac{du}{dx}=2x\), so \(du=2x\,dx\). The integral contains \(4x\,dx\), which is twice that differential: \(4x\,dx=2\,du\).
Check:
The factor \(2\) carried into the integral is essential. It combines with the \(\frac14\) from integrating \(u^3\), leaving the coefficient \(\frac12\).
A Common Factor-Counting Error
A common mistake is to replace \(dx\) by \(du\) even when differentiating the substitution gives a constant other than \(1\). For example, with \(u=5x\), we have \(du=5\,dx\), so \(dx=\frac15du\). Writing \(dx=du\) drops a factor of \(\frac15\).
For a concrete case, consider \(\int\cos(5x)\,dx\). The incorrect calculation is
Returning to \(x\) would give \(\sin(5x)+C\), but differentiating that answer gives \(5\cos(5x)\), not \(\cos(5x)\). The corrected substitution includes the changed differential:
Indeed, \(\frac{d}{dx}\bigl(\frac15\sin(5x)\bigr)=\frac15\cos(5x)\cdot 5=\cos(5x)\). This check catches the missing factor immediately.
When to Be Careful About Rewriting \(dx\)
The notation \(dx=\frac{1}{g'(x)}\,du\) is useful only where \(g'(x)\ne 0\), because division by \(g'(x)\) is not defined at a point where that derivative is zero. Many routine substitutions avoid this issue by matching a factor \(g'(x)\,dx\) directly rather than isolating \(dx\).
The substitution still has to account for every factor in the integrand. If a proposed \(u=g(x)\) leaves other \(x\)-dependence that cannot be expressed in terms of \(u\), the integrand has not yet been fully transformed. Do not remove or duplicate a factor merely to make the integral look familiar. Write the differential equality, match the factor present, and check the final antiderivative by differentiating it.
Check Your Understanding
For each question, pay particular attention to the factor attached to the differential.
- If \(u=2x^2+1\), what is \(du\), and what multiple of \(du\) is \(x\,dx\)?
- When \(u=7x+4\), what expression should replace \(dx\) in terms of \(du\)?
- In \(\int 10x(5x^2+3)^2\,dx\), how does \(10x\,dx\) compare with \(du\) for \(u=5x^2+3\)?
- Why does \(\int e^{3x}\,dx\) require a constant factor when \(u=3x\)?
- What differentiation check would reveal a missing constant after substitution?
