Integrals of Exponential Functions

The indefinite integral of the natural exponential function is \[\displaystyle \int e^x\,dx=e^x+C\] For any base $a>0$, with $a\neq 1$, the corresponding formula is \[\displaystyle \int a^x\,dx=\frac{a^x}{\ln a}+C\] where $C$ is an arbitrary constant of integration. If the exponent is a linear expression of the form $kx+b$, with $k\neq 0$, the formula includes an additional factor of $1/k$: \[\displaystyle \int a^{kx+b}\,dx=\frac{a^{kx+b}}{k\ln a}+C\]

To find the indefinite integral of an exponential function $a^x$, where $a>0$ and $a\neq1$, I use the following formula:

$$ \int a^x\,dx=\frac{a^x}{\ln a}+C $$

Here, $\ln a$ is the natural logarithm of the base, while $C$ is an arbitrary real constant of integration.

Why does this formula work? The key is the derivative of an exponential function:

$$ \frac{d}{dx}a^x=a^x\ln a $$

Differentiating $a^x$ introduces a factor of $\ln a$. To reverse this process and find an antiderivative, I divide $a^x$ by that constant.

I can check the result by differentiating:

$$ \frac{d}{dx}\left(\frac{a^x}{\ln a}\right) =\frac{a^x\ln a}{\ln a} =a^x $$

The derivative is exactly the original function, confirming that the integration formula is correct.

For example, suppose I want to integrate $2^x$: $$ \int 2^x\,dx $$ Since the base is $a=2$, I simply substitute this value into the formula: $$ \int 2^x\,dx=\frac{2^x}{\ln 2}+C $$

The Special Case of the Natural Exponential Function

Integration becomes even simpler when the base is Euler's number $e$. Since $\ln e=1$, the logarithmic factor in the denominator disappears:

$$ \int e^x\,dx=e^x+C $$

To see why, I substitute $a=e$ into the general formula $ \int a^x\,dx=\frac{a^x}{\ln a}+C $ and simplify:

$$ \int e^x\,dx=\frac{e^x}{\ln e}+C = \frac{e^x}{1}+C = e^x+C $$

In other words, the natural exponential function is its own antiderivative. This is because differentiating $e^x$ gives $e^x$ again.

For example, if the integrand includes a constant factor, I can take it outside the integral using the constant multiple rule: $$ \int 5e^x\,dx=5\int e^x\,dx=5e^x+C $$

Integrating Exponential Functions with Linear Exponents
Integrating Exponential Functions with General Exponents

Integrating Exponential Functions with Linear Exponents

So far, the exponent has been $x$. But what happens when the exponent is a linear expression such as $kx+b$?

Consider an exponential function of the form $a^{kx+b}$, where $k$ and $b$ are constants and $k\neq0$.

The integration formula becomes:

$$ \int a^{kx+b}\,dx=\frac{a^{kx+b}}{k\ln a}+C $$

The extra factor of $1/k$ is needed because of the chain rule. When I differentiate $a^{kx+b}$, the derivative of the exponent introduces a factor of $k$, in addition to the factor $\ln a$.

Dividing by $k$ accounts for this additional factor.

When the base is $e$, the formula simplifies to:

$$ \int e^{kx+b}\,dx=\frac{e^{kx+b}}{k}+C $$

For example, consider the following integral: $$ \int e^{3x+2}\,dx $$ The exponent is $3x+2$, and its derivative is $3$. To account for this factor, I divide the exponential expression by $3$: $$ \int e^{3x+2}\,dx=\frac{e^{3x+2}}{3}+C $$ I can verify the result by differentiating: $$ \frac{d}{dx}\left(\frac{e^{3x+2}}{3}\right) =\frac{3e^{3x+2}}{3} =e^{3x+2} $$

Integrating Exponential Functions with General Exponents

When the integrand contains an exponential function $e^{f(x)}$ multiplied by the derivative of its exponent $f'(x)$, I can apply the following rule: $$ \int f'(x)e^{f(x)}\,dx=e^{f(x)}+C $$ For any base $a>0$, with $a\neq1$, the corresponding formula is: $$ \int f'(x)a^{f(x)}\,dx=\frac{a^{f(x)}}{\ln a}+C $$

These formulas follow directly from the chain rule for differentiation. When integrating, I can use the same relationship in reverse through a technique called u-substitution.

The important point is to recognize the derivative of the exponent in the integrand.

For example, consider:

$$ \int 2x\,e^{x^2}\,dx $$

The exponent is $f(x)=x^2$, and its derivative is $f'(x)=2x$.

Since the factor $2x$ is already present in the integrand, I can apply the formula directly:

$$ \int 2x\,e^{x^2}\,dx=e^{x^2}+C $$

Note. To apply these formulas directly, the integrand must contain the derivative of the exponent as a factor, possibly multiplied by a nonzero constant that can be factored out. For example, consider $\int e^{x^2}\,dx$. This integral looks similar to the previous one, but the factor $2x$ is missing. As a result, the same integration rule cannot be applied directly. In fact, its antiderivative cannot be expressed in terms of elementary functions.

Now consider a similar integral with an arbitrary base:

$$ \int 2x\,a^{x^2}\,dx $$

Once again, the exponent is $x^2$, and its derivative is $2x$. Since this derivative appears in the integrand, I can use the same approach.

First, I make the substitution $u=x^2$. Differentiating gives $du=2x\,dx$, so the original integral becomes an integral of $a^u$.

The general rule behind this substitution is:

$$ \int u'(x)\,a^{u(x)}\,dx = \frac{a^{u(x)}}{\ln a} + C $$

Applying the substitution $u=x^2$ and $du=2x\,dx$, I obtain:

$$ \int 2x\,a^{x^2}\,dx =\int a^u\,du =\frac{a^u}{\ln a}+C =\frac{a^{x^2}}{\ln a}+C $$

This formula is valid for $a>0$ and $a\neq1$. If $a=1$, the exponential factor becomes $1$, so the integrand reduces to $2x$. Its indefinite integral is therefore $x^2+C$.

The same method works for many other exponential integrals. The key is to examine the exponent and check whether its derivative appears elsewhere in the integrand. Recognizing this relationship often turns a seemingly complicated integral into a straightforward application of the integration formula.