Question

Use any basic integration formula or formulas to find the indefinite integral. State which integration formula(s) you used to find the integral. $$ \int \frac{2}{1+e^{-x}} d x $$

Short answer

The indefinite integral of the function \(\frac{2}{1+e^{-x}}\) is \(2(e^{x} + 1) + C\). The basic integration formula used was the basic power rule for integration.

Step-by-step solution

Identifying the Function and Applying the Integral Formula

The function \(\frac{2}{1+e^{-x}}\) might seem a bit complex at first sight. We need to manipulate it a bit for it to be in a form we can integrate. It's important to know that \(e^{-x}\) is the same as \(\frac{1}{e^{x}}\) so we rewrite the function as \(\frac{2}{1+\frac{1}{e^x}}\), which can be further simplified to \(\frac{2e^{x}}{e^{x}+1}\). Now we see that we can make a \(u\) substitution. Set \(u = e^{x} +1\), then \(du = e^{x} dx\), so the integral becomes \(\int \frac{2u}{u} du\). This simplifies to \(\int 2 du\).

Performing the Integration

Now that the integral is simplified, we can perform the integration. The integral of \(2\) with respect to \(u\) is \(2u + C\), where \(C\) is the constant of integration.

Substitution back

Finally, we substitute \(u\) back in terms of \(x\), to get our result in terms of \(x\). So we obtain \(2u + C = 2(e^{x} + 1) + C\).

Key concepts

Integration by Substitution

Integration by substitution is a powerful technique that helps to solve complex integrals by simplifying them into a more manageable form. The key idea is to identify a part of the integral that can be substituted with a new variable, usually denoted as \( u \). This is often a component of the integrand that stands out or makes the expression easier to handle when replaced.

Here's how the process typically works:

• Identify a portion of the function to rewrite in terms of \( u \). In our problem, we set \( u = e^{x} + 1 \).
• Calculate the differential \( du \), in this case, \( du = e^{x} \, dx \).
• Replace the original variables and differential in the integral with \( u \) and \( du \). This changes the integral into a simpler form.
• Once the integral is solved, substitute back to express the answer in terms of the original variable \( x \).

This method is particularly useful when dealing with composite functions where one might represent a derivative of another part.

Exponential Function

An exponential function is a mathematical expression in which a constant base is raised to a variable exponent. The function \( e^{x} \) is a common example, where \( e \) is a constant approximately equal to 2.71828. Exponential functions are crucial in calculus and appear in many integrals and derivatives.

In the given exercise, the exponential function \( e^{-x} \) inside the integral significantly influences how the problem is solved. Notice how the original function \( \frac{2}{1+e^{-x}} \) required a rewriting to facilitate integration. Understanding the properties of exponential functions, including how they behave under transformation (like \( e^{-x} = \frac{1}{e^x} \)), is essential.

Exponential functions are known for their growth rates, and differentiating or integrating them results in other exponential functions, often with some transformation. The ability to manipulate and rewrite these accurately is key to solving integrals like this one.

Basic Integration Formulas

Basic integration formulas are fundamental rules used to find antiderivatives of functions. These formulas simplify the integration process by providing a direct path to the integral of standard functions.

In our problem, the basic integration formula applied was for a constant function because, after substitution, the integral simplified to \( \int 2 \, du \). The corresponding integral formula \( \int a \, dx = ax + C \) (where \( a \) is a constant) was used. This leads to the result \( 2u + C \).

• Power Rule: \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \)
• Exponential Rule: \( \int e^x \, dx = e^x + C \)
• Constant Multiple Rule: \( \int af(x) \, dx = a \int f(x) \, dx \)

Understanding these basic formulas aids in recognizing patterns and knowing how to manipulate expressions to fit these common forms. They form the backbone of solving many integral problems, including those involving substitution or exponential functions.