Chapter 1: Problem 10
Find the antiderivative of \(\int \frac{d x}{16+x^{2}}\).
Short Answer
Expert verified
The antiderivative is \( \frac{1}{4} \tan^{-1} \left( \frac{x}{4} \right) + C \).
Step by step solution
01
Recognize the Integral Form
The integral given is \( \int \frac{dx}{16 + x^2} \). This is a standard form of the antiderivative that matches the arctangent derivative formula: \( \int \frac{dx}{a^2 + x^2} = \frac{1}{a} \tan^{-1} \left(\frac{x}{a}\right) + C \), where \( a = 4 \).
02
Identify the Constant \(a\)
In the standard formula for \( \int \frac{dx}{a^2 + x^2} \), we need to identify \( a \). Comparing \( 16 + x^2 \) to \( a^2 + x^2 \), we see that \( a^2 = 16 \). Thus, \( a = 4 \).
03
Apply the Antiderivative Formula
Substitute \( a = 4 \) into the antiderivative formula to get \( \int \frac{dx}{16 + x^2} = \frac{1}{4} \tan^{-1} \left(\frac{x}{4}\right) + C \). This is the solution using substitution for the standard form.
04
Write the Final Answer
Thus, the antiderivative of the given integral is \( \frac{1}{4} \tan^{-1} \left( \frac{x}{4} \right) + C \). Remember that \( C \) represents the constant of integration.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Antiderivatives
In calculus, an antiderivative is essentially the reverse process of differentiation. Think of it as finding a function whose derivative is the given function. If you have the function's derivative, you need the antiderivative to get back to the original function. The antiderivative is not unique; instead, it includes a constant of integration, often denoted as \( C \). This is because when you differentiate a constant, it becomes zero, so any constant could have been part of the original function. Let's take a closer look at the standard antiderivative example used in our exercise. For a function like \( \int \frac{dx}{a^2 + x^2} \), the antiderivative is \( \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C \). - Recognize the role of the constant \( a \) in the formula.- Understand that \( C \) accounts for an unknown constant.Whenever you solve an integral problem like this, make sure to identify constants and apply the antiderivative formula correctly.
Integration Techniques
Integration techniques are methods used to solve integrals—functions under the integral sign. Various techniques can help tackle integrals of different forms, such as substitution, integration by parts, and trigonometric integrals. Knowing what strategy to use is crucial in solving integrals efficiently.In our exercise, a specific technique is applied: recognizing and using standard forms. - **Recognize standard forms:** Certain integrals follow known patterns. For example, \( \int \frac{dx}{a^2 + x^2} = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C \) fits a standard arctangent integral form because it matches the pattern where \( a \) is a constant.- **Substitution:** Identify constants in the equation. For the given integral, compare \( 16 + x^2 \) with \( a^2 + x^2 \) to determine \( a \).These techniques streamline the integration process by comparing to known integral solutions. Understanding when these methods apply can save significant time and simplify complex integrals.
Arctangent Function
The arctangent function, denoted as \( \tan^{-1}(x) \), is the inverse of the tangent function. It gives the angle whose tangent is a specific value. In calculus, the derivative of \( \tan^{-1}(x) \) is \( \frac{1}{1 + x^2} \), which relates directly to integration involving forms like \( \int \frac{dx}{a^2 + x^2} \).- **Inverse Trigonometric Functions:** Arctangent is one of these functions, playing a vital role in integration. - **Application in Integration:** Knowing that \( \int \frac{dx}{a^2 + x^2} = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C \), you realize that the integral itself is derived from the inverse trigonometric identity. Finding standard forms like this arctangent integral simplifies solving integrations involving quadratic expressions.When dealing with integrals involving quadratic denominators, recognizing the arctangent form is key, and it shows how inverse functions aid in the integration process.