Problem 59 - Calculus for Biology and Medicine

Question

Evaluate each integral. $$ \int \frac{1}{x^{2}+9} d x $$

Step-by-Step Solution

Answer

The integral evaluates to \( \frac{1}{3} \tan^{-1}\left( \frac{x}{3} \right) + C \).

Step 1: Recognize the Integral Form

The given integral \( \int \frac{1}{x^{2}+9} \, dx \) resembles the standard form \( \int \frac{1}{u^{2}+a^{2}} \, du \), which integrates to \( \frac{1}{a} \tan^{-1}\left( \frac{u}{a} \right) + C \).

Step 2: Identify Parameters

In the integral \( \int \frac{1}{x^{2}+9} \, dx \), we can identify \( u = x \) and \( a = 3 \), since \( 9 \) is the square of \( 3 \).

Step 3: Apply Standard Integral Formula

Using the standard formula \( \int \frac{1}{u^{2}+a^{2}} \, du = \frac{1}{a} \tan^{-1}\left( \frac{u}{a} \right) + C \), substitute \( a = 3 \) and \( u = x \). The integral then becomes \( \int \frac{1}{x^{2}+9} \, dx = \frac{1}{3} \tan^{-1}\left( \frac{x}{3} \right) + C \).

Step 4: Conclude with the Solution

Thus, the integral \( \int \frac{1}{x^{2}+9} \, dx \) evaluates to \( \frac{1}{3} \tan^{-1}\left( \frac{x}{3} \right) + C \), where \( C \) is the constant of integration.

Key Concepts

Definite Integrals

Definite integrals represent the signed area under a curve within a given interval on the x-axis. Unlike indefinite integrals, which provide a family of functions, definite integrals calculate a specific numerical value. The notation for definite integrals is \[\int_{a}^{b} f(x) \, dx,\]where \(a\) and \(b\) are the lower and upper bounds, respectively. To find the definite integral, we first find the antiderivative of the function \(f(x)\), then calculate its values at \(x = b\) and \(x = a\), and finally take the difference:

• Evaluate the antiderivative at \(b\).
• Evaluate the antiderivative at \(a\).
• Subtract the value at \(a\) from the value at \(b\).

This subtraction step gives the accumulated area between the curve and the x-axis from \(a\) to \(b\), effectively measuring the net signed area.

Inverse Trigonometric Functions

Inverse trigonometric functions provide the angle whose trigonometric function yields a given number. For example, if \(y = \tan^{-1}(x)\), then \(x = \tan(y)\). Among the most commonly used inverse trigonometric functions in calculus are \(\sin^{-1}(x)\), \(\cos^{-1}(x)\), and \(\tan^{-1}(x)\).These functions are crucial in integration, especially when you encounter integrals like \[\int \frac{1}{x^2 + a^2} \, dx,\]which results in an expression involving \(\tan^{-1}(x)\). The standard result is:\[\frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C,\]where \(a\) is a constant and \(C\) is the integration constant. Understanding the properties of inverse trigonometric functions allows you to handle integrals that appear in various calculus problems.

Integration Techniques

Integration techniques are essential tools for evaluating integrals that cannot be directly applied using basic rules. Here are some common techniques:

• **Substitution**: Useful when the integral can be simplified by changing variables.
• **Integration by Parts**: Essential for products of functions, based on the product rule of differentiation.
• **Partial Fractions**: Used for rational functions, it involves breaking down complex fractions into simpler ones.
• **Standard Integrals**: Some forms, like \(\int \frac{1}{x^2 + a^2} \, dx\), match known integrals linked to inverse functions.

In the exercise at hand, identifying the integral's form as a standard formula involving inverse trigonometric functions allows us to solve it efficiently. This efficient handling relies on recognizing patterns that fit these standard forms, thus making the process smoother and quicker.