Introduction
An exponential function with base \(e\) has a particularly simple antiderivative: it is its own antiderivative, up to an additive constant. For another positive base, such as \(2\), \(5\), or \(\frac{1}{3}\), the antiderivative still has a simple form, but it includes a factor that depends on the base.
That factor is essential. Differentiating \(a^x\) produces not only \(a^x\), but also a factor of \(\ln a\). The antiderivative must compensate for that factor. This tutorial derives the formula, states exactly when it applies, and checks it in several cases.
The Formula and Its Conditions
Here \(a\) is a fixed real number called the base, and \(x\) is the variable. The notation \(\int a^x\,dx\) asks for all functions whose derivative with respect to \(x\) is \(a^x\). The constant \(C\) appears because adding a constant to a function does not change its derivative.
The conditions on \(a\) matter. A positive base allows \(a^x\) to be defined as a real-valued function for every real \(x\). The base must also differ from \(1\), because \(\ln 1=0\), and division by \(\ln a\) would then be undefined. The special case \(a=1\) is easy to handle separately: \(1^x=1\), whose antiderivatives are \(x+C\).
For \(a>1\), \(\ln a\) is positive. For \(0<a<1\), \(\ln a\) is negative. The same formula works in both cases; the sign of the denominator automatically accounts for the behavior of the function.
Why the Logarithm of the Base Appears
For positive \(a\), the exponential function with base \(a\) can be written using the natural exponential: \(a^x=e^{x\ln a}\). This identity is the key connection. The natural logarithm of the base becomes a factor when the exponent \(x\ln a\) is differentiated.
Using the chain rule and the derivative of the natural exponential, we obtain $$ \frac{d}{dx}a^x = \frac{d}{dx}e^{x\ln a} = e^{x\ln a}\frac{d}{dx}(x\ln a) = e^{x\ln a}\ln a = (\ln a)a^x. $$
In this calculation \(a\) is constant, so \(\ln a\) is also constant and the derivative of \(x\ln a\) is \(\ln a\). Thus the derivative of \(a^x\) contains an extra multiplier \(\ln a\). To obtain \(a^x\) as the derivative of an antiderivative, divide by that multiplier.
Deriving and Checking the Antiderivative
For a fixed \(a>0\), use \(a^x=e^{x\ln a}\).
The chain rule gives \(\frac{d}{dx}a^x=(\ln a)a^x\).
When \(a\ne 1\), \(\ln a\ne0\), so differentiating \(\frac{a^x}{\ln a}\) cancels the factor \(\ln a\).
Indeed, the proposed function satisfies $$ \frac{d}{dx}\left(\frac{a^x}{\ln a}\right) = \frac{1}{\ln a}\frac{d}{dx}a^x = \frac{1}{\ln a}(\ln a)a^x = a^x. $$ This verifies directly that \(\frac{a^x}{\ln a}\) is an antiderivative.
To see why the arbitrary constant gives all antiderivatives on the real line, suppose \(F'(x)=a^x\) and let \(G(x)=a^x/\ln a\). The calculation above gives \(G'(x)=a^x\), so \((F-G)'(x)=0\) for every real \(x\). By the mean value theorem, a differentiable function on an interval whose derivative is zero everywhere has the same value at any two points in that interval: for points \(u<v\), the theorem gives a point \(c\in(u,v)\) such that \((F-G)(v)-(F-G)(u)=(F-G)'(c)(v-u)=0\). Since the real line is an interval, \(F-G\) is constant. Therefore every antiderivative has the form \(a^x/\ln a+C\).
Worked Examples
Worked Example: Find an Antiderivative of \(2^x\)
Here the base is \(a=2\). It satisfies \(2>0\) and \(2\ne1\), so the formula applies: $$ \int 2^x\,dx=\frac{2^x}{\ln 2}+C. $$ To check, differentiate the proposed antiderivative. Since \(\ln 2\) is a nonzero constant, $$ \frac{d}{dx}\left(\frac{2^x}{\ln 2}+C\right) = \frac{1}{\ln 2}\frac{d}{dx}(2^x) = \frac{1}{\ln 2}(\ln 2)2^x = 2^x. $$ The derivative is the original integrand, so the answer is correct.
Worked Example: Find an Antiderivative When the Base Is Between Zero and One
Find an antiderivative of \(\left(\frac{1}{5}\right)^x\). The base \(\frac{1}{5}\) is positive and is not \(1\). The formula gives $$ \int \left(\frac{1}{5}\right)^x\,dx = \frac{\left(\frac{1}{5}\right)^x}{\ln\left(\frac{1}{5}\right)}+C. $$ Because \(\ln\left(\frac{1}{5}\right)=-\ln 5\), the same answer can be written $$ -\frac{\left(\frac{1}{5}\right)^x}{\ln 5}+C. $$ Check by differentiating this equivalent form: $$ \frac{d}{dx}\left(-\frac{\left(\frac{1}{5}\right)^x}{\ln 5}+C\right) = -\frac{1}{\ln 5}\left(\ln\left(\frac{1}{5}\right)\right)\left(\frac{1}{5}\right)^x = -\frac{1}{\ln 5}(-\ln 5)\left(\frac{1}{5}\right)^x = \left(\frac{1}{5}\right)^x. $$ The negative sign in the antiderivative is necessary: the logarithm of a base between zero and one is negative.
Worked Example: Use a Given Function Value to Determine the Constant
Suppose \(F'(x)=3^x\) and \(F(0)=2\). First write the general antiderivative: $$ F(x)=\frac{3^x}{\ln 3}+C. $$ Now substitute \(x=0\) and use \(3^0=1\): $$ 2=F(0)=\frac{3^0}{\ln 3}+C=\frac{1}{\ln 3}+C. $$ Solving for \(C\) gives $$ C=2-\frac{1}{\ln 3}. $$ Thus the particular function is $$ F(x)=\frac{3^x}{\ln 3}+2-\frac{1}{\ln 3}. $$ Check both requirements. Differentiating, $$ F'(x)=\frac{1}{\ln 3}(\ln 3)3^x+0=3^x. $$ Evaluating at zero, $$ F(0)=\frac{1}{\ln 3}+2-\frac{1}{\ln 3}=2. $$ Both the derivative and the specified function value agree with the conditions.
A Common Mistake: Omitting \(\ln a\)
A frequent error is to assume that every exponential function is its own antiderivative. For example, the incorrect calculation $$ \int 2^x\,dx=2^x+C $$ would imply that the derivative of \(2^x\) is \(2^x\). But differentiating shows $$ \frac{d}{dx}(2^x+C)=(\ln 2)2^x, $$ which is not \(2^x\), because \(\ln 2\ne1\). The correction is to divide by \(\ln 2\): $$ \int 2^x\,dx=\frac{2^x}{\ln 2}+C. $$ Differentiating this corrected expression gives \(\frac{1}{\ln 2}(\ln 2)2^x=2^x\).
This error is a useful reminder to check an antiderivative by differentiating it. In the general case, the factor \(\ln a\) is not optional: it is exactly the factor generated by the chain rule.
Interpreting the Sign of the Formula
When \(a>1\), both \(a^x\) and \(\ln a\) are positive, so the formula \(a^x/\ln a\) is positive. When \(0<a<1\), \(a^x\) is positive but \(\ln a\) is negative, so the displayed antiderivative is negative. That negative value is not an error. Its derivative is positive, as the check in the second example confirms.
The case \(a=1\) also explains why the condition \(a\ne1\) must be stated. Substituting \(a=1\) into the general formula would require division by \(\ln 1=0\), which is undefined. Instead, integrate the constant function directly: $$ \int 1^x\,dx=\int 1\,dx=x+C. $$
Summary
For a fixed positive base other than one, the antiderivative of \(a^x\) is obtained by dividing by the natural logarithm of the base: $$ \int a^x\,dx=\frac{a^x}{\ln a}+C, \qquad a>0,\quad a\ne1. $$ The reason is that differentiating \(a^x\) produces \((\ln a)a^x\). Dividing by \(\ln a\) cancels that extra factor, and differentiating the proposed antiderivative confirms the result. When applying the formula, check the base conditions, keep the logarithm of the base in the denominator, and verify the answer by differentiation.
Check Your Understanding
Use the formula and, where requested, check your answer by differentiating.
- Find an antiderivative of \(7^x\).
- Find an antiderivative of \(\left(\frac{1}{4}\right)^x\), and explain the sign of the denominator in the formula.
- Differentiate \(\frac{5^x}{\ln 5}\). What does the result tell you?
- Why can the formula not be used with \(a=1\), and what is an antiderivative of \(1^x\)?
- Suppose \(G'(x)=2^x\) and \(G(0)=0\). Determine \(G(x)\).
