5.6 Integrals Involving Exponential and Logarithmic Functions

Exponential and logarithmic functions show up constantly in integration problems. Their integrals follow clean, predictable patterns, and once you know the core formulas and when to apply substitution, these problems become very manageable.

Integration Techniques for Exponential and Logarithmic Functions

Integrals of exponential functions

The natural exponential function is its own antiderivative, which makes it one of the simplest functions to integrate:

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

For exponential functions with a base other than $e$, you divide by the natural log of the base:

$\int a^x \, dx = \frac{a^x}{\ln a} + C, \quad a > 0, \; a \neq 1$

For example, $\int 2^x \, dx = \frac{2^x}{\ln 2} + C$ and $\int 10^x \, dx = \frac{10^x}{\ln 10} + C$.

Constants pull out front as usual:

• $\int k \cdot e^x \, dx = k \cdot e^x + C$
• $\int k \cdot a^x \, dx = \frac{k \cdot a^x}{\ln a} + C$

When an exponential is multiplied by $x$, you need integration by parts (which you may or may not have covered yet depending on your course). The results are:

• $\int x \cdot e^x \, dx = (x - 1) \cdot e^x + C$
• $\int x \cdot a^x \, dx = \frac{x \cdot a^x}{\ln a} - \frac{a^x}{(\ln a)^2} + C$

Integration of logarithmic expressions

You can't integrate $\ln x$ by a simple formula the way you can with $e^x$. Instead, the result comes from integration by parts, but the formula itself is worth memorizing:

$\int \ln x \, dx = x \ln x - x + C$

Constants factor out as expected:

• $\int k \cdot \ln x \, dx = k(x \ln x - x) + C$

For logarithms with a base other than $e$, use the change of base identity $\log_a x = \frac{\ln x}{\ln a}$ to convert, then integrate:

$\int \log_a x \, dx = \frac{x \ln x - x}{\ln a} + C, \quad a > 0, \; a \neq 1$

When a logarithm is multiplied by $x$, integration by parts gives:

• $\int x \cdot \ln x \, dx = \frac{x^2}{2} \ln x - \frac{x^2}{4} + C$

Substitution for exponential and logarithmic integrals

Many exponential and logarithmic integrals aren't in their basic form. That's where $u$-substitution comes in. The key idea: let $u$ equal whatever is "inside" the function (the exponent, or the argument of the logarithm), then rewrite everything in terms of $u$.

Example 1: Exponential substitution

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

1. Let $u = 2x$, so $du = 2 \, dx$, which means $dx = \frac{du}{2}$.
2. Rewrite: $\int e^{2x} \, dx = \int e^u \cdot \frac{du}{2} = \frac{1}{2} \int e^u \, du$.
3. Integrate and substitute back: $\frac{1}{2} e^u + C = \frac{1}{2} e^{2x} + C$.

Example 2: Logarithmic substitution

Evaluate $\int \ln(3x) \, dx$.

1. Let $u = 3x$, so $du = 3 \, dx$, which means $dx = \frac{du}{3}$.
2. Rewrite: $\int \ln(3x) \, dx = \int \ln u \cdot \frac{du}{3} = \frac{1}{3} \int \ln u \, du$.
3. Integrate and substitute back: $\frac{1}{3}(u \ln u - u) + C = \frac{1}{3}(3x \ln(3x) - 3x) + C$.

This simplifies to $x \ln(3x) - x + C$.

Compositions that simplify directly: Watch for cases where exponential and logarithmic functions cancel each other out. Since $e^{\ln x} = x$, you get:

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

Fundamental Concepts of Integration

A few background ideas tie everything together:

• An indefinite integral represents a family of antiderivatives, all differing by a constant $C$.
• A definite integral calculates the net area under a curve between two bounds.
• The Fundamental Theorem of Calculus connects differentiation and integration: if $F$ is an antiderivative of $f$, then $\int_a^b f(x) \, dx = F(b) - F(a)$.
• The exponential function $e^x$ and the natural logarithm $\ln x$ are inverses of each other. This relationship is why compositions like $e^{\ln x} = x$ and $\ln(e^x) = x$ simplify so cleanly, and it's the reason these two function types keep appearing together in integration problems.