1.6 Integrals Involving Exponential and Logarithmic Functions

Integration Techniques for Exponential and Logarithmic Functions

Exponential and logarithmic functions show up constantly in calculus, from modeling population growth to radioactive decay to compound interest. Knowing how to integrate them fluently is essential for the rest of Calculus II, especially when you hit differential equations and series later on.

Integration of Exponential Functions

The most important exponential integral to know is the natural exponential function $e^x$, because it's its own antiderivative. Everything else builds from there.

Core formulas:

• $\int e^x \, dx = e^x + C$
• $\int e^{ax} \, dx = \frac{1}{a} e^{ax} + C$ (where $a$ is a nonzero constant)
• $\int a^x \, dx = \frac{a^x}{\ln a} + C$ (where $a > 0$ and $a \neq 1$)

The second formula comes from a quick u-substitution: let $u = ax$, so $du = a\,dx$, and the $\frac{1}{a}$ falls out naturally. The third formula works because the derivative of $a^x$ is $a^x \ln a$, so you divide by $\ln a$ to reverse it.

Example 1: $\int e^{3x}\,dx$

Let $u = 3x$, so $du = 3\,dx$, meaning $dx = \frac{du}{3}$.

$\int e^{3x}\,dx = \frac{1}{3}\int e^u\,du = \frac{1}{3}e^{3x} + C$

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

This requires integration by parts (not just a formula to memorize). Set $u = x$ and $dv = e^{2x}\,dx$, so $du = dx$ and $v = \frac{1}{2}e^{2x}$.

$\int x\,e^{2x}\,dx = \frac{x}{2}e^{2x} - \int \frac{1}{2}e^{2x}\,dx = \frac{x}{2}e^{2x} - \frac{1}{4}e^{2x} + C = \frac{1}{4}(2x - 1)e^{2x} + C$

For integrals of the form $\int x^n e^{ax}\,dx$ with $n \geq 2$, you apply integration by parts repeatedly, reducing the power of $x$ by one each time.

Integrals with Logarithmic Functions

The connection between logarithms and integration starts with one key fact: the derivative of $\ln x$ is $\frac{1}{x}$. Reversing that gives you the most important formula in this section.

Core formulas:

• $\int \frac{1}{x}\,dx = \ln|x| + C$
• $\int \ln x\,dx = x\ln x - x + C$

The absolute value in the first formula matters. Since $\ln x$ is only defined for $x > 0$, but $\frac{1}{x}$ exists for all $x \neq 0$, the absolute value extends the antiderivative to negative $x$ values as well.

The second formula comes from integration by parts. Here's how:

1. Set $u = \ln x$ and $dv = dx$
2. Then $du = \frac{1}{x}\,dx$ and $v = x$
3. Apply the parts formula: $\int \ln x\,dx = x\ln x - \int x \cdot \frac{1}{x}\,dx = x\ln x - x + C$

Example 1: $\int \ln(2x)\,dx$

You can use parts the same way (set $u = \ln(2x)$, $dv = dx$), and you'll get $x\ln(2x) - x + C$.

Example 2: $\int \frac{1}{3x}\,dx$

Pull the constant out: $\frac{1}{3}\int \frac{1}{x}\,dx = \frac{1}{3}\ln|x| + C$.

Logarithms with other bases: If you need to integrate $\log_a x$, use the change of base formula $\log_a x = \frac{\ln x}{\ln a}$, then integrate $\frac{1}{\ln a}\int \ln x\,dx$ using the formula above.

Substitution for Exponential and Logarithmic Integrals

U-substitution is your main tool when the exponent or the argument of a logarithm is more complicated than just $x$. The key idea: look for a function paired with its derivative inside the integrand.

Steps for u-substitution:

1. Identify an inner function $g(x)$ whose derivative $g'(x)$ also appears in the integrand
2. Let $u = g(x)$, then compute $du = g'(x)\,dx$
3. Rewrite the entire integral in terms of $u$ and $du$
4. Integrate with respect to $u$
5. Substitute back to express the result in terms of $x$

Exponential example: $\int 2x\,e^{x^2}\,dx$

The exponent is $x^2$, and its derivative $2x$ sits right there in the integrand.

1. Let $u = x^2$, so $du = 2x\,dx$
2. The integral becomes $\int e^u\,du = e^u + C$
3. Substitute back: $e^{x^2} + C$

Logarithmic example: $\int \frac{1}{x\ln x}\,dx$

Here the integrand has the form $\frac{g'(x)}{g(x)}$ with $g(x) = \ln x$.

1. Let $u = \ln x$, so $du = \frac{1}{x}\,dx$
2. The integral becomes $\int \frac{1}{u}\,du = \ln|u| + C$
3. Substitute back: $\ln|\ln x| + C$

Applications in Differential Equations and Inverse Functions

These integration techniques become essential tools in later topics. Separable differential equations, for instance, often produce integrals of the form $\int \frac{1}{y}\,dy = \ln|y| + C$ or $\int e^{kt}\,dt = \frac{1}{k}e^{kt} + C$ when modeling exponential growth and decay.

Logarithmic integration also appears when finding antiderivatives of inverse trigonometric and other inverse functions, since integration by parts on these functions follows the same strategy as $\int \ln x\,dx$ (set the inverse function as $u$ and $dv = dx$).