Integral of the Exponential Function

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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:

\frac{d}{dx}\alpha^x = \alpha^x \ln \alpha

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:

\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 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 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:

\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 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 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 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. 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