Why Substitution Works Well with Exponentials
The derivative of a composite exponential has a predictable form: differentiate the exponential, then multiply by the derivative of its exponent. This means that an integral containing both an exponential expression and the derivative of its exponent can often be recognized as a reverse chain rule.
For example, differentiating \(e^{x^2}\) gives \(2x e^{x^2}\). Therefore, an integral containing \(2x e^{x^2}\) has exactly the ingredients needed to reverse that differentiation. Substitution provides a systematic way to identify the relationship rather than relying on pattern recognition alone.
The Exponential Substitution Rule
Let \(g\) be differentiable on an interval. The chain rule gives \(\frac{d}{dx}e^{g(x)}=e^{g(x)}g'(x)\). Reversing this derivative gives the basic integration rule.
The factor \(g'(x)\) is essential: it is what accounts for the chain rule when the proposed antiderivative is differentiated. If the integrand contains a constant multiple \(c g'(x)\), the constant simply passes through: \(\int e^{g(x)}c g'(x)\,dx=c\,e^{g(x)}+C\). A compensating factor is needed in the opposite situation, when \(g'(x)=c\,h(x)\) for a nonzero constant \(c\) and the integrand contains only \(h(x)\): then \(\int e^{g(x)}h(x)\,dx=\frac{1}{c}e^{g(x)}+C\).
This adjustment follows because the derivative of \(\frac{1}{c}e^{g(x)}\) is \(\frac{1}{c}e^{g(x)}g'(x)=\frac{1}{c}e^{g(x)}\,c\,h(x)=e^{g(x)}h(x)\). In practice, it is safer to write the substitution explicitly and solve for the differential.
Deriving the Rule from Substitution
The rule is a direct consequence of substitution and the antiderivative of \(e^u\). The derivation also shows why matching the inner derivative matters.
Set \(u=g(x)\), so that \(du=g'(x)\,dx\).
The product \(e^{g(x)}g'(x)\,dx\) becomes \(e^u\,du\).
Since \(\int e^u\,du=e^u+C\), the integral becomes \(e^u+C\).
Substitute \(g(x)\) for \(u\), obtaining \(e^{g(x)}+C\).
Thus the substitution does not create a new integration formula. It converts a composite expression into the familiar exponential antiderivative, provided the differential factor matches.
Worked Examples with Base \(e\)
Worked Example: A Polynomial Exponent with Its Derivative
Evaluate \(\int 6x e^{3x^2-4}\,dx\).
The inner expression is \(3x^2-4\), whose derivative is \(6x\). Set \(u=3x^2-4\). Then \(du=6x\,dx\).
Substitute both the exponential and the matching differential:
Check by differentiation: \(\frac{d}{dx}e^{3x^2-4}=e^{3x^2-4}(6x)=6x e^{3x^2-4}\), which is the integrand. Therefore the antiderivative is correct.
Worked Example: Adjusting for a Missing Constant Factor
Evaluate \(\int 4x e^{2x^2+1}\,dx\).
Let \(u=2x^2+1\). Then \(du=4x\,dx\). The differential in the integral matches exactly, so no extra factor is required:
Check: \(\frac{d}{dx}e^{2x^2+1}=e^{2x^2+1}(4x)=4x e^{2x^2+1}\).
Now compare with \(\int 2x e^{2x^2+1}\,dx\). The same inner expression has derivative \(4x\), but the integral contains only \(2x\). Since \(du=4x\,dx\), it follows that \(2x\,dx=\frac{1}{2}du\). Thus
Check: \(\frac{d}{dx}\left(\frac{1}{2}e^{2x^2+1}\right)=\frac{1}{2}e^{2x^2+1}(4x)=2x e^{2x^2+1}\), as required.
Exponentials with a Base Other Than \(e\)
For a positive constant base \(a\ne 1\), the derivative rule is \(\frac{d}{du}a^u=a^u\ln a\). The extra factor \(\ln a\) must be included when integrating. Consequently, \(\int a^u\,du=\frac{a^u}{\ln a}+C\). This gives a substitution rule for a composite exponent as well.
Worked Example: Integrating a Composite Exponential with Base 5
Evaluate \(\int 5^{2x+3}\,dx\).
The exponent is \(2x+3\), with derivative \(2\), while the integrand has no visible factor of \(2\). Set \(u=2x+3\). Then \(du=2\,dx\), so \(dx=\frac{1}{2}du\). Substitute:
Check by differentiation:
The factor \(\ln 5\) from differentiating the exponential cancels with the denominator, and the factor \(2\) from the exponent cancels with the other denominator.
Worked Example: Separating Two Constant Adjustments
Worked Example: A Constant Outside and a Scaled Inner Derivative
Evaluate \(\int 12e^{4x-1}\,dx\).
Let \(u=4x-1\), so \(du=4\,dx\) and \(dx=\frac{1}{4}du\). Keeping the outside coefficient \(12\) visible,
Check: \(\frac{d}{dx}\left(3e^{4x-1}\right)=3e^{4x-1}(4)=12e^{4x-1}\). The original coefficient and the derivative of the exponent combine to give the required factor.
Recognizing the Pattern and Its Limits
A practical way to inspect an exponential integral is to identify the exponent first, then differentiate it. Compare that derivative with the factors multiplying the exponential. The match may be exact or may differ by a constant.
| Integrand structure | What to notice |
|---|---|
| \(e^{g(x)}g'(x)\) | The derivative of the exponent is present exactly. |
| \(e^{g(x)}c g'(x)\), with constant \(c\ne0\) | Use \(g'(x)\,dx=du\); the constant \(c\) multiplies the result. |
| \(e^{g(x)}h(x)\), where \(g'(x)=c\,h(x)\) with constant \(c\ne0\) | Use \(h(x)\,dx=\frac{1}{c}du\), so divide by \(c\). |
| \(a^{g(x)}g'(x)\), where \(a>0\) and \(a\ne1\) | Use the antiderivative factor \(\frac{1}{\ln a}\). |
| An exponential with no matching derivative factor | This substitution pattern does not directly apply as written. |
For instance, in \(\int e^{x^2}\,dx\), choosing \(u=x^2\) gives \(du=2x\,dx\), but the integral has no factor of \(x\). The substitution therefore does not turn the integral into a constant multiple of \(\int e^u\,du\). This does not justify inserting a missing factor; it means this particular direct pattern is unavailable.
Common Mistake: Multiplying by the Derivative Instead of Adjusting for It
A frequent error is to notice the derivative of the exponent and multiply by it again, even though the integral already contains a different factor. For example, consider \(\int 2x e^{2x^2+1}\,dx\). The exponent’s derivative is \(4x\), so the integrand contains half of that derivative.
A wrong calculation might say, “The exponent differentiates to \(4x\), so the answer is \(4e^{2x^2+1}\).” But differentiating that proposed answer gives \(\frac{d}{dx}(4e^{2x^2+1})=16x e^{2x^2+1}\), not \(2x e^{2x^2+1}\). The coefficient is incorrect.
The correct substitution is \(u=2x^2+1\), \(du=4x\,dx\), and therefore \(2x\,dx=\frac{1}{2}du\). It follows that \(\int 2x e^{2x^2+1}\,dx=\frac{1}{2}e^{2x^2+1}+C\). Differentiating this result gives \(2x e^{2x^2+1}\), exactly the original integrand.
Summary
For exponential substitution, the exponent is the natural choice for the new variable. If \(u=g(x)\), then \(du=g'(x)\,dx\), so the integral simplifies when the integrand contains that derivative, possibly up to a constant factor. For base \(e\), the antiderivative of \(e^u\) is \(e^u\); for base \(a>0\), \(a\ne1\), the antiderivative is \(\frac{a^u}{\ln a}\).
The central check is always the chain rule in reverse: differentiate the proposed antiderivative and confirm that every coefficient and exponential factor reproduces the original integrand.
Check Your Understanding
For each question, identify the inner expression and account for any constant factor. Check each proposed antiderivative by differentiation.
- Evaluate \(\int 8x e^{4x^2-2}\,dx\).
- Evaluate \(\int 3x^2 e^{x^3+5}\,dx\).
- Evaluate \(\int e^{6x-1}\,dx\).
- Evaluate \(\int 2^{3x+2}\,dx\), stating the role of \(\ln 2\).
- Why does the direct exponential substitution pattern not immediately evaluate \(\int e^{x^2}\,dx\)?
