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Antiderivatives and Basic Integrals · Tutorial 1067 of 1495

Exponential Integrals With Constant Coefficients

Use the exponential antiderivative formulas and linearity to integrate sums while keeping every coefficient and logarithmic factor correct.

Beginner 8 min read

What You'll Learn

  • State when the general-base exponential antiderivative formula applies
  • Integrate constant multiples of natural exponential functions
  • Keep the logarithm of each base attached to the correct term
  • Apply linearity to finite sums of exponential functions
  • Verify an exponential antiderivative by differentiating it
  • Recognize and correct a missing logarithmic factor

Why the Coefficients Matter

Exponential functions have a characteristic derivative: differentiating an exponential function produces the same exponential function multiplied by a constant. Integration reverses that multiplication. As a result, an exponential antiderivative often needs a factor that is easy to overlook.

For the natural exponential \(e^x\), that factor is \(1\), because \(\ln e=1\). For an exponential with a different base, the factor involves the logarithm of that base. When several exponential terms appear in a sum, each term must keep its own coefficient and its own logarithmic factor.

Central rule: For a positive base \(a\ne 1\), the antiderivative of \(a^x\) is \(\frac{a^x}{\ln a}+C\). The natural exponential is the special case \(a=e\), for which \(\ln e=1\).

The Antiderivative Rules

An antiderivative of a function \(f\) on an interval is a function \(F\) whose derivative equals \(f\) throughout that interval: \(F'(x)=f(x)\). The indefinite integral records the family of antiderivatives, which differ by an additive constant on an interval.

The exponential formulas below apply for all real \(x\). For the general-base formula, the base must satisfy \(a>0\) and \(a\ne 1\). Positivity makes \(\ln a\) defined, while \(a\ne 1\) ensures that \(\ln a\ne 0\).

$$ \int e^x\,dx=e^x+C \qquad\text{and}\qquad \int a^x\,dx=\frac{a^x}{\ln a}+C \quad (a>0,\ a\ne 1). $$

Here \(C\) is an arbitrary constant. Since adding a constant does not change a derivative, every value of \(C\) gives an antiderivative. On the real line, which is one interval, all antiderivatives of the same function differ by a single constant.

Linearity allows constant multiples and finite sums to be handled term by term. If \(c_1,\ldots,c_n\) are constants and each \(f_i\) has an antiderivative \(F_i\) on the interval under consideration, then the following is an antiderivative of the sum:

$$ \int \left(c_1f_1(x)+\cdots+c_nf_n(x)\right)\,dx = c_1F_1(x)+\cdots+c_nF_n(x)+C. $$

In this tutorial, the terms \(f_i\) are natural or general-base exponential functions. The coefficient outside a term stays outside its antiderivative, while the factor \(1/\ln a\) belongs to the term with base \(a\). There is only one final \(+C\) for the sum.

Why the General-Base Formula Works

The base-change identity writes a general-base exponential in terms of the natural exponential: \(a^x=e^{x\ln a}\). Differentiating this expression gives \(\frac{d}{dx}a^x=a^x\ln a\). The logarithmic factor in the antiderivative formula reverses that multiplication.

Derivation: For \(a>0\) with \(a\ne1\), differentiate the proposed antiderivative. The constant \(1/\ln a\) is well-defined, and the chain rule gives
$$ \frac{d}{dx}\left(\frac{a^x}{\ln a}\right) = \frac{1}{\ln a}\frac{d}{dx}\left(e^{x\ln a}\right) = \frac{1}{\ln a}e^{x\ln a}\ln a = a^x. $$
Thus \(\frac{a^x}{\ln a}\) is an antiderivative of \(a^x\). Adding any constant \(C\) leaves its derivative unchanged.

For the natural exponential, substituting \(a=e\) gives \(\ln e=1\), so the general formula becomes \(e^x/\ln e=e^x\). This explains why no visible logarithmic denominator appears for \(e^x\).

For a finite sum, the justification is just as direct. Differentiate the proposed sum of antiderivatives. The derivative of each term is its corresponding integrand term, with its constant coefficient unchanged. Adding those derivatives recovers the original sum.

Worked Examples

Worked Example: A Natural Exponential and a General Base

Find an antiderivative of \(4e^x-3\cdot2^x\).

Integrate each term separately. The coefficient \(4\) remains with \(e^x\). For the base \(2\), divide by \(\ln 2\):

$$ \begin{aligned} \int (4e^x-3\cdot2^x)\,dx &=4\int e^x\,dx-3\int 2^x\,dx\\ &=4e^x-\frac{3\cdot2^x}{\ln 2}+C. \end{aligned} $$

Check by differentiation. Since \(\frac{d}{dx}2^x=2^x\ln 2\),

$$ \frac{d}{dx}\left(4e^x-\frac{3\cdot2^x}{\ln 2}+C\right) = 4e^x-\frac{3}{\ln 2}(2^x\ln 2) = 4e^x-3\cdot2^x. $$

The derivative matches the integrand, so the antiderivative is correct.

Worked Example: Keeping a Separate Logarithm for Each Base

Find an antiderivative of \(2\cdot5^x+7e^x-4\left(\frac12\right)^x\).

Use the natural exponential rule for the middle term. For each other term, divide by the logarithm of that term’s base:

$$ \begin{aligned} \int \left(2\cdot5^x+7e^x-4\left(\frac12\right)^x\right)\,dx &= \frac{2\cdot5^x}{\ln 5} +7e^x -\frac{4\left(\frac12\right)^x}{\ln\left(\frac12\right)} +C. \end{aligned} $$

Because \(\ln(1/2)=-\ln 2\), the last term can also be written with a positive sign:

$$ \frac{2\cdot5^x}{\ln 5} +7e^x +\frac{4\left(\frac12\right)^x}{\ln 2} +C. $$

Check the result in this equivalent form. Differentiating each term gives

$$ \begin{aligned} \frac{d}{dx}\left( \frac{2\cdot5^x}{\ln 5} +7e^x +\frac{4\left(\frac12\right)^x}{\ln 2} +C \right) &= 2\cdot5^x+7e^x +\frac{4\left(\frac12\right)^x\ln(1/2)}{\ln 2}\\ &= 2\cdot5^x+7e^x-4\left(\frac12\right)^x. \end{aligned} $$

The negative sign in the last term of the derivative comes from \(\ln(1/2)=-\ln 2\). It agrees with the original integrand.

Worked Example: Integrating a Sum with Fractional Coefficients

Find an antiderivative of \(\frac{1}{3}e^x+\frac{5}{2}\cdot3^x\).

Apply the constant-multiple rule to each term. The first term uses the natural exponential formula; the second uses the base \(3\) formula:

$$ \begin{aligned} \int \left(\frac{1}{3}e^x+\frac{5}{2}\cdot3^x\right)\,dx &= \frac{1}{3}\int e^x\,dx +\frac{5}{2}\int 3^x\,dx\\ &= \frac{1}{3}e^x+\frac{5\cdot3^x}{2\ln 3}+C. \end{aligned} $$

Differentiate to check:

$$ \begin{aligned} \frac{d}{dx}\left( \frac{1}{3}e^x+\frac{5\cdot3^x}{2\ln 3}+C \right) &= \frac{1}{3}e^x+\frac{5}{2\ln 3}(3^x\ln 3)\\ &= \frac{1}{3}e^x+\frac{5}{2}\cdot3^x. \end{aligned} $$

Both the fractional coefficient and the logarithmic factor are necessary for the derivative to recover the given term.

A Reliable Process for a Sum

When integrating a finite sum of constant multiples of exponentials, keep the work organized term by term. The following process makes it easier to see which base determines each logarithmic factor.

1
Identify each term and its base.
Separate the constant coefficient from the exponential function. A term involving \(e^x\) uses the natural exponential rule; a term involving \(a^x\) uses the general-base rule.
2
Check the base conditions.
For a general base, confirm \(a>0\) and \(a\ne1\), so \(\ln a\) exists and is nonzero.
3
Integrate each term.
Keep its constant coefficient, and divide its exponential by the natural logarithm of its own base when the base is not \(e\).
4
Add one integration constant and check.
Write one \(+C\) for the whole sum, then differentiate the result to see whether it returns every original term with the correct sign and coefficient.

Common Mistake: Forgetting the Logarithm of the Base

A frequent error is to use \(a^x\) itself as an antiderivative of \(a^x\) for every base. That is only correct when \(a=e\). For example, the incorrect calculation \(\int 3^x\,dx=3^x+C\) fails the derivative check:

$$ \frac{d}{dx}(3^x+C)=3^x\ln 3\ne3^x. $$

The correction is to include the reciprocal logarithm of the base:

$$ \int 3^x\,dx=\frac{3^x}{\ln 3}+C, \qquad \frac{d}{dx}\left(\frac{3^x}{\ln 3}+C\right)=3^x. $$

In a sum, another version of this mistake is to use the logarithm of one base for every exponential term. Each general-base term has its own denominator. For example, \(2^x\) requires \(\ln 2\), while \(5^x\) requires \(\ln 5\). The safest check is to differentiate each term separately.

Summary

For \(a>0\) and \(a\ne1\), the antiderivative of \(a^x\) is \(a^x/\ln a+C\). The natural exponential is the special case \(e^x\), whose antiderivative is \(e^x+C\). For a finite sum, integrate each term separately, preserve its coefficient, use the logarithm of its own base, and include one arbitrary constant for the sum.

Differentiation provides a direct check: the logarithmic factor in each general-base antiderivative cancels the factor introduced when differentiating that exponential. If the derivative does not reproduce the original sum exactly, check the coefficient, sign, base, and logarithmic factor term by term.

Check Your Understanding

For each question, keep track of the coefficient and the base, and use differentiation to check your result.

  1. Find an antiderivative of \(6e^x+2\cdot4^x\).
  2. What conditions must a general base \(a\) satisfy for the formula \(\int a^x\,dx=a^x/\ln a+C\) to apply?
  3. Find an antiderivative of \(3\cdot2^x-5\cdot7^x\).
  4. Explain why \(a^x+C\) is generally not an antiderivative of \(a^x\) when \(a\ne e\).
  5. When integrating a finite sum of exponential terms, how many arbitrary constants are needed in the final expression?

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