Problem 44 - Calculus for Biology and Medicine

Question

In Problems \(41-46\), assume that \(a\) is a positive constant. Find the general antiderivative of the given function. $$ f(x)=\frac{a}{a+x} $$

Step-by-Step Solution

Answer

The general antiderivative is \( a\ln|a+x| + C \).

Step 1: Understand the Problem

We are tasked with finding the general antiderivative of the function \( f(x) = \frac{a}{a+x} \). An antiderivative of a function is another function whose derivative is the original function.

Step 2: Set Up the Integral

The general antiderivative of a function \( f(x) \) is found by integrating \( f(x) \). Thus, we need to compute \( \int \frac{a}{a+x} \, dx \).

Step 3: Apply Substitution

To solve \( \int \frac{a}{a+x} \, dx \), use the substitution method. Let \( u = a + x \), which gives \( du = dx \). The integral becomes \( \int \frac{a}{u} \, du \).

Step 4: Integrate Using the Constant Multiple Rule

Recognize that \( \frac{a}{u} \) is a constant \( a \) times \( \frac{1}{u} \). Integrate \( \int \frac{1}{u} \, du = \ln|u| + C \). Thus, \( \int \frac{a}{u} \, du = a \ln|u| + C \).

Step 5: Substitute Back Original Variable

Replace \( u \) with \( a + x \) to express the integral in terms of the original variable. This gives the antiderivative as \( a\ln|a+x| + C \).

Step 6: Final Antiderivative Form

The general antiderivative of \( f(x) = \frac{a}{a+x} \) is \( F(x) = a\ln|a+x| + C \), where \( C \) is the constant of integration.

Key Concepts

Understanding the Antiderivative

When tackling calculus problems, especially those involving antiderivatives, it's important to understand the basics first. An antiderivative of a function is a new function whose derivative equals the original function. It's the reverse of differentiation. For example, if the derivative of a certain function is given to you, finding the antiderivative means determining the original function.

This idea is crucial because an antiderivative allows us to solve a variety of calculus problems, including those dealing with rates of change and areas under curves. Keep in mind that there are an infinite number of antiderivatives since adding any constant to an antiderivative still satisfies the requirement of yielding the original function when differentiated.

Integration by Substitution

Integration by substitution is a handy technique for solving integrals that seem complex at first glance.

Here's how it generally works:

• We choose a part of the integral to replace with a single variable, usually denoted as \(u\). This transforms the integral into a simpler form.
• Next, we calculate the derivative of \(u\), known as \(du\). This will replace \(dx\) (or whatever differential is present in the integral).
• After substituting \(u\) into the integral, and replacing \(dx\) with \(du\), the integral becomes easier to solve.
• Finally, after integrating, substitute back to the original variable to express the result in terms of the initial function.

In our problem, we saw the transformation of \(\int \frac{a}{a+x} \, dx\) by letting \(u = a + x\), which simplifies it to \(\int \frac{a}{u} \, du\). This change made the integral easier to manage and solve.

The Constant of Integration

When finding antiderivatives, you're often working with indefinite integrals, which have a family of solutions. This is where the constant of integration, typically denoted as \(C\), comes into play.

• Adding \(C\) accounts for the fact that there are multiple functions whose derivative can yield the given function.
• The constant \(C\) can take any real number value. Thus, the antiderivative isn't a single function but rather a family of functions.
• This constant is particularly crucial in real-world applications, like setting initial conditions in physics to pinpoint a specific solution.

In the exercise, after performing the integration, the constant \(C\) is added to form the general antiderivative: \(F(x) = a\ln|a+x| + C\).

Logarithmic Integration

Logarithmic integration comes into play with integrals resulting in logarithmic functions due to their forms. Many integrals that involve \(\frac{1}{x}\) or similar expressions have solutions rooted in the natural logarithm function, \(\ln\). For example:

• The integration of \(\frac{1}{x}\) leads directly to \(\ln|x| + C\) because the derivative of \(\ln|x|\) is \(\frac{1}{x}\).
• Recognizing patterns involving logarithmic forms simplifies more complex integrals.
• This understanding helps automate recognizing when to apply \(\ln\) as the antiderivative, especially during substitution.

In our problem, once we used substitution and simplified, the integral required integrating \(\frac{1}{u}\), leading to the logarithmic result of \(a\ln|a+x| + C\).