# Integral of the Exponential Function Source: algebrica.org - CC BY-NC 4.0 https://algebrica.org/integral-of-the-exponential-function/ Fetched from algebrica.org post 8073; source modified 2026-02-25T18:57:36. ## How the integral of the exponential function is calculated An [exponential function](../exponential-function.md) is a function of the form \\( e^x \\) or \\( \\alpha^x \\) (with \\( \\alpha > 0 \\) and \\( \\alpha \\neq 1 \\)). In general, the number \\( e \\) occupies a central position in analysis because it is the only base for which the exponential function reproduces itself under differentiation. For a general exponential function \\( \\alpha^x \\) with \\( \\alpha > 0 \\), differentiation introduces an unavoidable factor: \\[ \\frac{d}{dx}\\alpha^x = \\alpha^x \\ln \\alpha \\] That [logarithmic](../logarithms.md) term reflects how the chosen base scales the growth of the function. There is exactly one case in which this extra factor disappears. If \\( \\ln \\alpha = 1 \\), we have: \\[ \\frac{d}{dx}\\alpha^x = \\alpha^x \\] The unique number satisfying this condition is \\( e \\approx 2.718 \\). Thus \\( e^x \\) is the only exponential function that remains unchanged by differentiation. The same simplicity extends to integration. This structural property explains why \\( e \\) plays such a fundamental role. * * * Knowing how to compute the [integral](../indefinite-integrals.md) of such functions is very useful in exercises involving exponential terms. We have two cases. The integral of \\( e^x \\) is given by: \\[ \\int e^x \\,dx = e^x + c \\tag{1} \\] Indeed, it can be proven that the [derivative](../derivatives.md) of \\( e^x \\) is itself \\( e^x \\). By differentiating the integral result: \\[ D\[e^x + c\] = D\[e^x\] + D\[c\] = e^x + 0 = e^x \\] ##### Delve into how Euler’s number \\(e\\) can be defined as the [limit of a known sequence](../euler-number-limit-sequence.md). * * * The integral of \\(\\alpha^x\\) is given by: \\[ \\int \\alpha^x \\,dx =\\frac{1}{\\ln\\alpha} \\cdot \\alpha^x + c \\tag{2} \\] In fact, we have: \\[ D \\left\[ \\frac{1}{\\ln \\alpha} \\cdot \\alpha^x + c \\right\] = \\frac{1}{\\ln \\alpha} \\cdot (\\ln \\alpha \\cdot \\alpha^x) = \\alpha^x \\] ## Canonical forms of exponential integrals ###### Each row displays a standard exponential integrand on the left and its corresponding antiderivative on the right. These canonical patterns encompass the forms most commonly encountered in integration problems. - \\[ \\text{1. } \\quad \\int e^x \\, dx \\] \\[ e^x + c \\] - \\[ \\text{2. } \\quad \\int \\alpha^x \\, dx \\] \\[ \\dfrac{1}{\\ln \\alpha}\\,\\alpha^x + c \\] - \\[ \\text{3. } \\quad \\int e^{ax+b} \\, dx \\] \\[ \\dfrac{1}{a} \\,e^{ax+b} + c \\] - \\[ \\text{4. } \\quad \\int \\alpha^{ax} \\, dx \\] \\[ \\dfrac{1}{a \\ln \\alpha}\\,\\alpha^{ax} + c \\] - \\[ \\text{5. } \\quad \\int e^{f(x)} f’(x) \\, dx \\] \\[ e^{f(x)} + c \\] Exponential functions preserve their structure under integration: the form remains unchanged, and only a multiplicative constant reflects the rate encoded in the exponent. ## Example 1 Let’s consider the following integral: \\[ \\int e^x + 3^x \\,dx \\] * * * By the [linearity property](../indefinite-integrals.md) of the integral, the integral of a sum is equal to the sum of the integrals: \\[ \\int (f(x) + g(x)) \\,dx = \\int f(x) \\,dx + \\int g(x) \\,dx \\] We have: \\[ \\int e^x + 3^x \\, dx = \\int e^x \\, dx + \\int 3^x \\, dx \\] * * * The first integral can be easily derived from \\( (1) \\): \\[ \\int e^x \\, dx = e^x + c \\] The second integral can be derived from \\( (2) \\): \\[ \\int 3^x \\, dx = \\frac{1}{\\ln3} \\cdot 3^x + c \\] Thus, our integral becomes: \\[ e^x + \\frac{1}{\\ln3} \\cdot 3^x + c \\] ## Exponential with a linear argument In practice, the exponent is rarely just \\(x\\). A very common situation is an exponential whose argument is a linear function \\(ax + b\\), with \\(a \\neq 0\\). In that case we have: \\[ \\int e^{ax+b} \\,dx = \\frac{1}{a} e^{ax+b} + c \\] The factor \\(\\frac{1}{a}\\) compensates exactly for what the [chain rule](../the-derivative-of-a-composite-function.md) introduces when differentiating. To verify: \\[ D \\! \\left\[\\frac{1}{a} e^{ax+b} + c\\right\] = \\frac{1}{a} \\cdot a \\cdot e^{ax+b} = e^{ax+b} \\] More generally, when the exponent is a differentiable function \\(f(x)\\), [integration by substitution](../integration-by-substitution.md) gives: \\[ \\int e^{f(x)} \\cdot f’(x) \\,dx = e^{f(x)} + c \\] If the integrand contains an exponential \\(e^{f(x)}\\) multiplied by the derivative of its own exponent, the integral collapses cleanly to \\(e^{f(x)} + c\\). When \\(f’(x)\\) is not present, an algebraic manipulation or substitution is needed first. * * * The same reasoning based on the chain rule applies to exponential functions with an arbitrary base \\( \\alpha \\). If the exponent is \\( ax \\) instead of just \\( x \\), differentiation produces two factors: the coefficient \\( a \\) from the exponent and \\( \\ln \\alpha \\) from the base. For this reason we have: \\[ \\int \\alpha^{ax} \\, dx = \\frac{1}{a \\ln \\alpha} \\, \\alpha^{ax} + c \\] A direct differentiation confirms the formula. This identity is useful in practice, since expressions of this type often appear in intermediate steps when simplifying more complicated integrals. ## Example 2 Let us now consider the following integral, which at first glance appears slightly more complex than the one presented in example 1. \\[ \\int 8^x \\cdot 2^{(-3x + 4)} \\, dx \\] * * * To solve it, we can take advantage of the [properties of powers](../powers.md). We can rewrite: \\[ 2^{-3x+4} = 2^{-3x} \\cdot 2^4 = 2^{-3x} \\cdot 16 \\] The integral then becomes: \\[ \\begin{aligned} 16 \\int 8^x \\cdot 2^{-3x} \\, dx &= 16 \\int (2^3)^x \\cdot 2^{-3x} \\, dx \\[6pt] &= 16 \\int 2^{3x} \\cdot 2^{-3x} \\, dx \\[6pt] &= 16 \\int 2^{3x-3x} \\, dx \\[6pt] &= 16 \\int 1 \\, dx \\end{aligned} \\] We obtain: \\[ 16x + c \\] ## Example 3 Let’s consider the following integral: \\[ \\int 9^{x-1} \\cdot 3^{-x+2} \\, dx \\] * * * We can rewrite the integral using the properties of powers: \\[ \\int 9^x \\cdot 9^{-1} \\cdot 3^{-x} \\cdot 3^2 \\, dx = \\int 9^x \\cdot 9^{-1} \\cdot 3^{-x} \\cdot 9 \\, dx \\] Simplifying the terms, we obtain: \\[ \\int 9^x \\cdot 3^{-x} \\, dx = \\int 3^{2x} \\cdot 3^{-x} \\, dx = \\int 3^{x} \\, dx \\] We have reduced the integral to the form: \\[ \\int \\alpha^x \\,dx \\] We obtain \\[ \\frac{1}{\\ln3} \\cdot 3^x + c \\] ## Example 4 Consider the following integral: \\[ \\int e^{3x - 2} \\, dx \\] The exponent is linear, so this is a direct application of the standard rule for exponential functions of the form \\( e^{ax+b} \\). Since the derivative of \\( 3x - 2 \\) is \\( 3 \\), we compensate by dividing by \\( 3 \\). \\[ \\int e^{3x-2} \\, dx = \\frac{1}{3} e^{3x-2} + c \\] It is always worth checking the result. Differentiating \\( \\frac{1}{3} e^{3x-2} \\) gives: \\[ \\frac{1}{3} \\cdot 3 e^{3x-2} = e^{3x-2} \\] so the computation is consistent. The solution is: \\[ \\frac{1}{3} e^{3x-2} + c \\] ## A common oversight A frequent source of error arises when the exponent carries a coefficient. It is easy to write: \\[ \\int e^{3x-2}\\,dx = e^{3x-2} + c \\] and overlook the factor \\( \\frac{1}{3} \\). The issue becomes clear as soon as one differentiates the result: \\[ \\frac{d}{dx} e^{3x-2} = 3 e^{3x-2} \\] which is not the original integrand but three times as large. The check takes only a moment and immediately reveals the inconsistency. Whenever the exponent has the form \\( ax + b \\) with \\( a \\neq 1 \\), the compensating factor \\( \\frac{1}{a} \\) is essential. ## Example 5 Consider now the following integral, in order to examine this new situation: \\[ \\int x \\, e^{x^2} \\, dx \\] Here the exponent is \\( x^2 \\), which is not linear, so the previous rule for \\( e^{ax+b} \\) cannot be applied directly. However, the structure of the integrand suggests what to do. The derivative of \\( x^2 \\) is \\( 2x \\), and a factor \\( x \\) is already present. We rewrite the integral by introducing the constant: \\[ \\int x \\, e^{x^2} \\, dx = \\frac{1}{2} \\int 2x \\, e^{x^2} \\, dx \\] Now the integrand has the form \\( e^{f(x)} f’(x) \\) with \\( f(x) = x^2 \\). In this situation, integration is immediate: \\[ \\frac{1}{2} e^{x^2} + c \\] A quick verification confirms the result. Differentiating \\( \\frac{1}{2} e^{x^2} \\) produces: \\[ \\frac{1}{2} \\cdot 2x \\, e^{x^2} = x e^{x^2} \\] which matches the original integrand. The solution is: \\[ \\frac{1}{2} e^{x^2} + c \\] ## When the matching factor is missing It is useful to notice what happens if the factor \\( x \\) is removed. The integral: \\[ \\int e^{x^2}\\,dx \\] does not have an antiderivative that can be written using elementary functions. Instead, it is expressed in terms of the error function \\( \\mathrm{erf}(x) \\). It is defined by the integral: \\[ \\mathrm{erf}(x) = \\frac{2}{\\sqrt{\\pi}} \\int\_0^x e^{-t^2} \\, dt \\] Unlike elementary functions such as polynomials, exponentials, or trigonometric functions, \\( \\mathrm{erf}(x) \\) is defined directly through an integral. It was introduced precisely because integrals of the form \\( \\int e^{-t^2} dt \\) cannot be expressed in closed elementary form. This shows that the factor \\( x \\) in the previous example provided, up to a constant, the derivative of the exponent \\( x^2 \\). Without that match, the simple structure disappears and the integral can no longer be handled with the same elementary tools. ## Selected references - **University of California, Davis – L. Kouba**. [Integration of Exponential Functions](https://www.math.ucdavis.edu/~kouba/CalcTwoDIRECTORY/expondirectory/Exponentials.html) - **MIT, G. Strang**. [Calculus – Chapter 6: Exponentials and Logarithms](https://ocw.mit.edu/ans7870/textbooks/Strang/Edited/Calculus/6.pdf) - **University of Wisconsin–Madison, S. Angenent**. [MATH 222 – Second Semester Calculus](https://people.math.wisc.edu/~angenent/222.2013s/UWMath222_2013s.pdf) - **University of Chicago, V. Jayaram**. [Proving the Non-Existence of Elementary Anti-Derivatives](http://math.uchicago.edu/~may/REU2017/REUPapers/Jayaram.pdf)