Problem 15 - Calculus for Biology and Medicine

Question

In Problems 1-16, evaluate each indefinite integral by making the given substitution. $$ \int \frac{3 x}{x+2} d x, \text { with } u=x+2 $$

Step-by-Step Solution

Answer

The integral evaluates to \( 3x + 6 - 6 \ln |x+2| + C \).

Step 1: Make the Substitution

We are given the substitution \( u = x + 2 \). From this substitution, we can find \( du \). Since \( u = x + 2 \), it follows that \( du = dx \). Also, we can express \( x \) in terms of \( u \): \( x = u - 2 \).

Step 2: Rewrite Integral in Terms of u

Rewrite the given integral \( \int \frac{3x}{x+2} \, dx \) using our substitution. Substitute \( x = u - 2 \) and \( dx = du \) into the integral: \( \int \frac{3(u-2)}{u} \, du \).

Step 3: Simplify the Integral

Simplify \( \int \frac{3(u-2)}{u} \, du \) by splitting the fraction: \( \int \left( 3 - \frac{6}{u} \right) \, du \). This is now split into two separate integrals: \( 3 \int 1 \, du - 6 \int \frac{1}{u} \, du \).

Step 4: Evaluate Each Part

Now, evaluate each integral separately. First, \( 3 \int 1 \, du = 3u \). Second, \( -6 \int \frac{1}{u} \, du = -6 \ln |u| \). Therefore, the solution is \( 3u - 6 \ln |u| + C \), where \( C \) is the constant of integration.

Step 5: Back Substitute for x

Replace \( u \) with \( x+2 \) in the solution to return to the original variable \( x \): \( 3(x+2) - 6 \ln |x+2| + C \). Simplify this expression to \( 3x + 6 - 6 \ln |x+2| + C \).

Key Concepts

Substitution Method

The substitution method is an essential technique used in solving indefinite integrals, particularly when an integral seems too complex to handle directly. It works by transforming a complicated integral into a simpler one. You achieve this by changing the variables. For our given problem, we set the substitution to be \( u = x + 2 \). This choice simplifies the integral significantly and allows us to isolate the variable into terms of \( u \) by expressing both \( x \) and \( dx \) in terms of \( u \).

This step of substituting is crucial because it transforms the integration variable and adjusts the integral's format. Here’s a simple rundown of the process:

• Choose your substitution, usually to reduce the integral to a known form.
• Express the original variables in terms of the new variable \( u \).
• Find \( du \) in terms of the original differential.

Once the substitution is complete, you can integrate with respect to \( u \) much more easily.

Integration Techniques

Integration techniques are the various strategies used to solve integrals, and they can vary from direct integration to more complex methods like substitution and integration by parts. In the context of this problem, we employ the substitution technique and then simplify the resulting expression to separate integrals that are straightforward.

• After substitution, our goal is often to simplify the expression, making it a sum of simpler integrals.
• Here, we apply the property \( \frac{a}{b} = \frac{a}{b} + \frac{c}{b} - \cdots \) to split up the fraction \( \frac{3(u-2)}{u} \) into two separate terms \( 3 - \frac{6}{u} \).

These steps allow us to handle the integrals individually, which then can be solved using simple rules of integration:

• \( \int 1 \, du = u \)
• \( \int \frac{1}{u} \, du = \ln |u| \)

With these simple rules, the integration process becomes systematic and straightforward.

Calculus Problem Solving

Effective calculus problem-solving involves understanding the structure of a problem and applying the appropriate methods to find a solution. This is not just about following procedures but understanding why these steps work:

• Firstly, identify the problem type and any substitutions that simplify it.
• Then, break the problem into smaller, more manageable parts.
• Finally, apply integration rules and back substitute to return to the original variable when needed.

In our integration example, following these steps ensures that we effectively simplify the integral from one in terms of \( x \) to one in terms of \( u \), which is easier to solve. We then evaluate each piece independently before back substituting to express the solution in terms of the original variable, \( x \). Understanding these problem-solving strategies in calculus allows for better comprehension of how to approach and solve a wide array of similar calculus problems.