Problem 54 - College Algebra and Calculus: An Applied Approach

Question

Use any basic integration formula or formulas to find the indefinite integral. State which integration formula(s) you used to find the integral. $$ \int \frac{5}{e^{-5 x}+7} d x $$

Step-by-Step Solution

Answer

The indefinite integral \(\int \frac{5}{e^{-5 x}+7} d x\) is equal to \(- \ln |e^{-5x}+7| + C\). The integration formula used here is the basic logarithmic integration rule, \(\int \frac{1}{u} du = \ln |u|\).

Step 1: Recognize the Integral Structure and Apply Change of Variable

The integral can be simplified by changing the variable. Let's define a new variable \(u\), such that \(u = e^{-5x}+7\). Therefore, \(du = -5e^{-5x} dx\). But we want the integral in terms of dx, hence we re-arrange to get \(dx = - \frac{1}{5 e^{-5x}} du \). The integral becomes \(\int \frac{5}{u} \cdot - \frac{1}{5 e^{-5x}} du.\)

Step 2: Simplication of Integral

The integral simplifies to \(- \int \frac{1}{u} du\). This is a standard integral, which can be solved with the logarithmic integration rule.

Step 3: Apply Logarithmic Integration Rule

Applying the logarithmic integration rule, which states that \(\int \frac{1}{u} du = \ln |u|\), to the integral \(- \int \frac{1}{u} du\) we get \(- \ln |u| + C\), where \(C\) is the constant of integration.

Step 4: Substitute Back the Original Variable

Our final step is to substitute \(u = e^{-5x}+7\) back into our solution. Thus, the solution to the original integral is \(- \ln |e^{-5x}+7| + C\).

Key Concepts

Integration by Substitution

Integration by substitution is a powerful technique in calculus. It allows us to simplify complex integrals by "substituting" a part of the integral with a new variable. This is often referred to as making a 'change of variable'. For example, if an integral involves a complicated expression like \(e^{-5x} + 7\), we can introduce a new variable, \(u\), and set \(u = e^{-5x} + 7\). This substitution helps to reframe the integral into a simpler form.
The goal of substitution is to transform the integral into one that is easier to solve. We also compute the differential \(du\) based on the new variable. In this case, differentiating both sides with respect to \(x\) gives \(du = -5e^{-5x}dx\). Next, we solve for \(dx\) to connect it back to \(du\), resulting in \(dx = - \frac{1}{5e^{-5x}} du\). This process effectively simplifies the integral by replacing complex expressions with \(u\), making our calculations more straightforward.

Logarithmic Integration Rule

The logarithmic integration rule is an essential tool when dealing with integrals of the form \(\int \frac{1}{u} du\). This rule simplifies these integrals to a natural logarithm, specifically \(\ln |u|\), plus a constant of integration \(C\). It's important to understand that this rule applies because the derivative of \(\ln |u|\) is \(\frac{1}{u}\). This makes it directly applicable whenever you arrive at an integral with \(\frac{1}{u}\) as the integrand.
In the example given, after substituting and simplifying, the integral becomes \(- \int \frac{1}{u} du\). Applying the logarithmic integration rule transforms it into \(- \ln |u| + C\). Remember that the \(-\) sign from the simplified integral carries through the calculation, emphasizing the importance of paying attention to signs when performing integration.

Change of Variables

The process of 'change of variables' involves substituting a part of the integrand with a new variable. This method is predominantly beneficial when the original variable section of the integration is overly complex. By doing this, it allows for a clearer path to take the integral of the expression.
Change of variables isn't limited to substitution; it can also include rewriting variables in a way that leverages existing mathematical rules more effectively. It's crucial to remember to switch back to the original variable once your integration is complete. For instance, after applying a change of variable \(u = e^{-5x} + 7\), and integrating with respect to \(u\), you'll eventually substitute \(u\) back in terms of \(x\). This ensures that the solution matches the context of the original problem.