AP Calculus AB and BC

Integral of a^x: Answer, Proof, and Mistakes

Q: What is the integral of a^x?

\int a^x\,dx = \frac{a^x}{\ln a} + C, for any constant base with a > 0 and a \neq 1. The base stays the same and picks up a denominator of \ln a.

Q: Why does the integral of a^x divide by ln a instead of multiplying?

Because the derivative multiplies. Since \frac{d}{dx}[a^x] = a^x \ln a, differentiating leaves an extra factor of \ln a, and the antiderivative has to divide it out so that the derivative of \frac{a^x}{\ln a} comes back to a^x. Checking with a = e settles it: only the dividing version returns e^x.

Q: What is the integral of 2^x?

\int 2^x\,dx = \frac{2^x}{\ln 2} + C. Numerically \frac{1}{\ln 2} \approx 1.4427, so the antiderivative is roughly 1.4427 \cdot 2^x.

Q: What is the integral of a^(kx)?

\int a^{kx}\,dx = \frac{a^{kx}}{k \ln a} + C. The substitution u = kx contributes the extra \frac{1}{k}, so you divide by the inside coefficient as well as by \ln a. For example, \int 7^{3x}\,dx = \frac{7^{3x}}{3 \ln 7} + C.

Q: How is the integral of a^x different from the integral of x^a?

They use opposite rules. In a^x the exponent is the variable, so the exponential rule applies and the answer is \frac{a^x}{\ln a} + C. In x^a the base is the variable, so the power rule applies and the answer is \frac{x^{a+1}}{a+1} + C for a \neq -1. Deciding which one you are looking at is the whole battle.

The integral of a^x is a^x divided by ln a, plus C, for any base a > 0 with a not equal to 1. So the integral of 2^x dx is 2^x/ln 2 + C. The rule reverses the derivative d/dx of a^x = a^x ln a: dividing by the constant ln a cancels the extra factor. When a = e, ln e = 1, so e^x integrates to itself.

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

How to prove the integral of a^x

Antidifferentiation is the reverse of differentiation, so the fastest route is to start from the derivative of $a^x$ and undo it. Rewriting $a^x = e^{x \ln a}$ and applying the chain rule, where $\ln a$ is a constant multiplier on the inside, gives the derivative rule.

\frac{d}{dx}\left[a^x\right] = a^x \ln a

Differentiating $a^x$ therefore returns the original function times an unwanted constant factor of $\ln a$ . Dividing by that constant before differentiating cancels it, since constant multiples pass straight through the derivative.

\frac{d}{dx}\left[\frac{a^x}{\ln a}\right] = \frac{a^x \ln a}{\ln a} = a^x

The derivative of $\frac{a^x}{\ln a}$ is exactly the integrand, so it is an antiderivative of $a^x$ . Every other antiderivative differs from it by a constant, which is what $+\,C$ records.

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

Why the base must be positive and not 1

The restriction $a > 0$ keeps $a^x$ defined for every real $x$ , and $a \neq 1$ keeps $\ln a \neq 0$ so the division is legal. If $a = 1$ the integrand is just the constant function 1, and $\int 1\,dx = x + C$ .

Where the integral of a^x shows up on the AP exam

Basic antiderivative rules like this one sit in Unit 6, Topic 6.8 (Finding Antiderivatives and Indefinite Integrals: Basic Rules and Notation). Unit 6 carries a weighting of 15 to 20 percent on both AB and BC, so it is one of the two most heavily weighted units on each exam.

The base $e$ is the case the College Board leans on hardest, because $\ln e = 1$ makes the denominator disappear. Reading the general rule with $a = e$ recovers the familiar result.

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

When the exponent is a linear function rather than bare $x$ , the integral needs substitution (Topic 6.9), and you divide by the inside coefficient as well as by $\ln a$ .

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

On the multiple-choice section this usually appears as a one-step recall question or as one term inside a longer sum. On free response it turns up inside accumulation and area setups, where an exponential model with a base other than $e$ has to be antidifferentiated before the Fundamental Theorem is applied.

Common mistakes with the integral of a^x

Almost every error on this rule is a sign error in the sense of doing the derivative operation instead of the integral one, or reaching for the power rule when the variable is upstairs.

Multiplying by $\ln a$ instead of dividing. Multiplication is the derivative, $a^x \ln a$ ; division is the integral, $\frac{a^x}{\ln a}$ . Sanity check with $a = e$ : only division gives back $e^x$ .
Applying the power rule to get $\frac{a^{x+1}}{x+1}$ . The power rule needs a constant exponent and a variable base. Here the base is constant and the exponent is the variable, which is the exponential rule instead.
Forgetting the inside coefficient on $a^{kx}$ . Writing $\int 5^{2x}\,dx = \frac{5^{2x}}{\ln 5}$ misses the factor of 2 from the substitution; the answer is $\frac{5^{2x}}{2 \ln 5} + C$ .
Writing $\ln a^x$ or $\ln(a)^x$ in the denominator. The denominator is the constant $\ln a$ , evaluated once, and it does not carry the variable exponent.
Dropping the constant of integration. An indefinite integral is a whole family of functions, and the missing $+\,C$ is a routine point loss on free response.

Worked examples

Example 1. A plain base swap. Apply the rule with $a = 3$ .

\int 3^x\,dx = \frac{3^x}{\ln 3} + C

Example 2. A linear exponent. Let $u = 2x$ , so $du = 2\,dx$ and $dx = \frac{du}{2}$ .

\int 5^{2x}\,dx = \frac{1}{2}\int 5^{u}\,du = \frac{5^{2x}}{2 \ln 5} + C

Example 3. A sum with the natural base alongside another base. Integrate term by term, and note that only the $2^x$ term picks up a logarithm in the denominator.

\int \left(4 \cdot 2^x - e^x\right)dx = \frac{4 \cdot 2^x}{\ln 2} - e^x + C

Example 4. A substitution with a nonlinear exponent. Let $u = x^2$ , so $du = 2x\,dx$ and the stray $x$ out front is exactly what the substitution needs.

\int x \cdot 3^{x^2}\,dx = \frac{1}{2}\int 3^{u}\,du = \frac{3^{x^2}}{2 \ln 3} + C

Example 5. A definite integral. Evaluate the antiderivative at the endpoints, and watch the $\ln 2$ survive as a constant in the exact answer.

\int_0^1 2^x\,dx = \left[\frac{2^x}{\ln 2}\right]_0^1 = \frac{2 - 1}{\ln 2} = \frac{1}{\ln 2} \approx 1.443

Every answer on this page is machine checked

An automated test differentiates the antiderivative above and confirms it returns the integrand. A wrong sign or a missing factor fails the build, so it cannot reach you. The base is a parameter, so the numeric check fixes a = 2: the integrand 2^x is cross-checked against the antiderivative 2^x/ln(2) at the sample points. The general result a^x/ln(a) + C is derived the same way for every a > 0 with a not equal to 1.

Frequently asked questions

What is the integral of a^x?

$\int a^x\,dx = \frac{a^x}{\ln a} + C$ , for any constant base with $a > 0$ and $a \neq 1$ . The base stays the same and picks up a denominator of $\ln a$ .

Why does the integral of a^x divide by ln a instead of multiplying?

Because the derivative multiplies. Since $\frac{d}{dx}[a^x] = a^x \ln a$ , differentiating leaves an extra factor of $\ln a$ , and the antiderivative has to divide it out so that the derivative of $\frac{a^x}{\ln a}$ comes back to $a^x$ . Checking with $a = e$ settles it: only the dividing version returns $e^x$ .

What is the integral of 2^x?

$\int 2^x\,dx = \frac{2^x}{\ln 2} + C$ . Numerically $\frac{1}{\ln 2} \approx 1.4427$ , so the antiderivative is roughly $1.4427 \cdot 2^x$ .

What is the integral of a^(kx)?

$\int a^{kx}\,dx = \frac{a^{kx}}{k \ln a} + C$ . The substitution $u = kx$ contributes the extra $\frac{1}{k}$ , so you divide by the inside coefficient as well as by $\ln a$ . For example, $\int 7^{3x}\,dx = \frac{7^{3x}}{3 \ln 7} + C$ .

How is the integral of a^x different from the integral of x^a?

They use opposite rules. In $a^x$ the exponent is the variable, so the exponential rule applies and the answer is $\frac{a^x}{\ln a} + C$ . In $x^a$ the base is the variable, so the power rule applies and the answer is $\frac{x^{a+1}}{a+1} + C$ for $a \neq -1$ . Deciding which one you are looking at is the whole battle.