Problem 18 - Calculus for Biology and Medicine

Question

Find the general antiderivative of the given function. $$ f(x)=\frac{1}{1+3 x} $$

Step-by-Step Solution

Answer

The general antiderivative is \( \frac{1}{3} \ln |1+3x| + C \).

Step 1: Understand the Function

We need to find the general antiderivative of the function \( f(x) = \frac{1}{1+3x} \). This function is a rational function where the numerator is constant and the denominator is linear. This suggests using logarithmic integration.

Step 2: Identify the Integral Formula

The function \( f(x) \) resembles \( \frac{1}{u} \), where \( u = 1 + 3x \). The antiderivative of \( \frac{1}{u} \) with respect to \( u \) is \( \ln |u| + C \), where \( C \) is an arbitrary constant.

Step 3: Apply Integration with Substitution

Let \( u = 1 + 3x \), then \( \frac{du}{dx} = 3 \), or \( du = 3 \, dx \). Therefore, \( dx = \frac{1}{3} \, du \). Substitute into the integral to get \( \int \frac{1}{u} \cdot \frac{1}{3} \, du \).

Step 4: Resolve the Integral

The integral becomes \( \frac{1}{3} \int \frac{1}{u} \, du = \frac{1}{3} \ln |u| + C \).

Step 5: Substitute Back in Terms of x

Substitute \( u = 1 + 3x \) back into the expression:\[ \int \frac{1}{1+3x} \, dx = \frac{1}{3} \ln |1 + 3x| + C \].

Step 6: Verify the Solution

Differentiate \( \frac{1}{3} \ln |1+3x| + C \) to confirm it's original function \( f(x) = \frac{1}{1+3x} \). The derivative \( \frac{d}{dx} \left( \frac{1}{3} \ln |1+3x| \right) = \frac{1}{1+3x} \). The differentiation verifies our integral is correct.

Key Concepts

Logarithmic Integration

Logarithmic integration occurs when you integrate functions of the form \( \frac{1}{u} \). This arises because the derivative of the natural logarithm, \( \ln |u| \), is \( \frac{1}{u} \). Therefore, the integral \( \int \frac{1}{u} \, du \) results in \( \ln |u| + C \), where \( C \) is an integration constant.
This rule is particularly useful for functions where you can set the expression in the denominator as a new variable \( u \).

When dealing with logarithmic integration:

• Identify the denominator, and consider it as a unit \( u \) that might simplify integration.
• Recognize when a simple \( \ln \) function represents the antiderivative.
• Remember to re-substitute back to the original variable after integrating.

Logarithmic integration makes it easy to handle rational functions, especially when the denominator is linear or of an easy-to-manage polynomial form.

Integration by Substitution

Integration by substitution, also known as the substitution rule, is a commonly used technique to simplify integrals.
This is especially useful when the integral seems complex at first glance, but can become easier by changing variables.

Here's how to apply integration by substitution:

• Identify a part of the integrand as \( u \) (preferably, the more complicated part). In our example, we set \( u = 1 + 3x \).
• Calculate \( \frac{du}{dx} \) and solve for \( dx \). In the solution, this gave \( dx = \frac{1}{3} \, du \).
• Substitute \( u \) and \( du \) into the original integral to obtain a simpler integral in terms of \( u \).
• Integrate with respect to \( u \).
• Finally, substitute back the expressions in terms of the original variable \( x \).

Using this method can transform difficult integrals into more manageable forms.

Rational Function

A rational function is a quotient of two polynomials. It is represented in the form \( \frac{P(x)}{Q(x)} \), where both \( P(x) \) and \( Q(x) \) are polynomials.
The properties of rational functions make them interesting entities to work with in calculus, particularly for integration.

When integrating rational functions, certain strategies come into play:

• Identify whether you can simplify the expression, such as by reducing or factoring polynomials.
• Use techniques like partial fraction decomposition if the denominator can be factored further into simpler linear or quadratic terms.
• Be alert for opportunities to apply logarithmic integration, especially if the denominator has a simple polynomial form.

In the example provided, the function was a basic rational function, making it a good candidate for logarithmic integration, as the denominator was linear in form and easily manageable.