Integral of the Exponential Function

How the integral of the exponential function is calculated

An exponential function 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:

That logarithmic 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:

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 of such functions is very useful in exercises involving exponential terms. We have two cases.

The integral of $e^{x}$ is given by: \((\text{1}) \int e^{x} d x = e^{x} + c\) Indeed, it can be proven that the derivative 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.

The integral of $\alpha^{x}$ is given by: \((\text{2}) \int \alpha^{x} d x = \frac{1}{ln ⁡ \alpha} \cdot \alpha^{x} + c\) In fact, we have: \(D [ \frac{1}{ln ⁡ \alpha} \cdot \alpha^{x} + c ] = \frac{1}{ln ⁡ \alpha} \cdot ( ln ⁡ \alpha \cdot \alpha^{x} ) = \alpha^{x}\)

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.

• \(\text{1}. \int e^{x} d x\) \(e^{x} + c\)
• \(\text{2}. \int \alpha^{x} d x\) \(\frac{1}{ln ⁡ \alpha} \alpha^{x} + c\)
• \(\text{3}. \int e^{a x + b} d x\) \(\frac{1}{a} e^{a x + b} + c\)
• \(\text{4}. \int \alpha^{a x} d x\) \(\frac{1}{a ln ⁡ \alpha} \alpha^{a x} + c\)
• \(\text{5}. \int e^{f ( x )} f^{'} ( x ) d x\) \(e^{f ( x )} + c\)

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: \(\int e^{x} + 3^{x} d x\)

By the linearity property of the integral, the integral of a sum is equal to the sum of the integrals:

We have:

The first integral can be easily derived from $( 1 )$: \(\int e^{x} d x = e^{x} + c\)

The second integral can be derived from $( 2 )$: \(\int 3^{x} d x = \frac{1}{ln ⁡ 3} \cdot 3^{x} + c\)

Thus, our integral becomes:

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 $a x + b$, with $a \neq 0$. In that case we have:

The factor $\frac{1}{a}$ compensates exactly for what the chain rule introduces when differentiating. To verify:

More generally, when the exponent is a differentiable function $f ( x )$, integration by substitution gives:

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 $a x$ 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:

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.

To solve it, we can take advantage of the properties of powers. We can rewrite:

The integral then becomes:

We obtain:

Example 3

Let’s consider the following integral: \(\int 9^{x - 1} \cdot 3^{- x + 2} d x\)

We can rewrite the integral using the properties of powers: \(\int 9^{x} \cdot 9^{- 1} \cdot 3^{- x} \cdot 3^{2} d x = \int 9^{x} \cdot 9^{- 1} \cdot 3^{- x} \cdot 9 d x\)

Simplifying the terms, we obtain: \(\int 9^{x} \cdot 3^{- x} d x = \int 3^{2 x} \cdot 3^{- x} d x = \int 3^{x} d x\)

We have reduced the integral to the form: \(\int \alpha^{x} d x\)

We obtain

Example 4

Consider the following integral:

The exponent is linear, so this is a direct application of the standard rule for exponential functions of the form $e^{a x + b}$. Since the derivative of $3 x - 2$ is $3$, we compensate by dividing by $3$.

It is always worth checking the result. Differentiating $\frac{1}{3} e^{3 x - 2}$ gives:

so the computation is consistent.

The solution is: \(\frac{1}{3} e^{3 x - 2} + c\)

A common oversight

A frequent source of error arises when the exponent carries a coefficient. It is easy to write:

and overlook the factor $\frac{1}{3}$. The issue becomes clear as soon as one differentiates the result:

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 $a x + b$ 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:

Here the exponent is $x^{2}$, which is not linear, so the previous rule for $e^{a x + b}$ cannot be applied directly. However, the structure of the integrand suggests what to do. The derivative of $x^{2}$ is $2 x$, and a factor $x$ is already present. We rewrite the integral by introducing the constant:

Now the integrand has the form $e^{f ( x )} f^{‘} ( x )$ with $f ( x ) = x^{2}$. In this situation, integration is immediate:

A quick verification confirms the result. Differentiating $\frac{1}{2} e^{x^{2}}$ produces:

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:

does not have an antiderivative that can be written using elementary functions. Instead, it is expressed in terms of the error function $erf ( x )$. It is defined by the integral:

Unlike elementary functions such as polynomials, exponentials, or trigonometric functions, $erf ( x )$ is defined directly through an integral. It was introduced precisely because integrals of the form $\int e^{- t^{2}} d t$ 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
• MIT, G. Strang. Calculus – Chapter 6: Exponentials and Logarithms
• University of Wisconsin–Madison, S. Angenent. MATH 222 – Second Semester Calculus
• University of Chicago, V. Jayaram. Proving the Non-Existence of Elementary Anti-Derivatives