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Magazine Chapter 28: Problem 28
Solve the given problems by integration. Perform the integration \(\int \frac{x d x}{x^{2}+1}\) (a) by using the logarithmic formula, and (b) by trigonometric substitution. Compare results.
Short Answer
Expert verified
Both integration methods give \( \frac{1}{2} \ln |x^2+1| + C \).
Step by step solution
01
- Integrate using the logarithmic formula
To integrate \( \int \frac{x}{x^2+1} \, dx \), we can recognize it as a logarithmic form. Let \( u = x^2 + 1 \). Then, \( du = 2x \, dx \) or \( \frac{1}{2} du = x \, dx \). Rewriting the integral, we have:\[\int \frac{x}{x^2+1} \, dx = \frac{1}{2} \int \frac{1}{u} \, du\]This simplifies to:\[\frac{1}{2} \ln |u| + C = \frac{1}{2} \ln |x^2 + 1| + C\]
02
- Integrate using trigonometric substitution
To solve \( \int \frac{x}{x^2+1} \, dx \) using trigonometric substitution, let \( x = \tan \theta \), then \( dx = \sec^2 \theta \, d\theta \), and \( x^2 + 1 = \tan^2 \theta + 1 = \sec^2 \theta \). Substituting, the integral becomes:\[\int \frac{\tan \theta \sec^2 \theta \, d\theta}{\sec^2 \theta} = \int \tan \theta \, d\theta\]The integral of \( \tan \theta \) is:\[-\ln |\cos \theta| + C\]Re-substituting \( \theta \) with \( \tan^{-1}(x) \), use \( \cos \theta = \frac{1}{\sqrt{x^2 + 1}} \) to get:\[-\ln \left| \frac{1}{\sqrt{x^2+1}} \right| = \frac{1}{2} \ln |x^2 + 1| + C\]
03
- Compare results
The integration using the logarithmic formula gives:\[\frac{1}{2} \ln |x^2+1| + C\]The integration using trigonometric substitution simplifies to the same form:\[\frac{1}{2} \ln |x^2+1| + C\]Both methods give the same result confirming the correctness of the integration methods.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Logarithmic Integration
Logarithmic integration is a handy technique especially when you encounter integrals of the form \( \int \frac{f'(x)}{f(x)} \, dx \). This type integrates to \( \ln |f(x)| + C \), where \( C \) is the constant of integration. In the example, we have the integral \( \int \frac{x}{x^2+1} \, dx \). Recognizing that the derivative of \( x^2 + 1 \) is \( 2x \), you can use the logarithmic rule here.
- Step 1: Set \( u = x^2 + 1 \). This means \( du = 2x \, dx \) or rearrange to get \( \frac{1}{2} du = x \, dx \).
- Step 2: Substitute into the integral to get: \( \frac{1}{2} \int \frac{1}{u} \, du \).
- Step 3: Integrate \( \frac{1}{u} \, du \), resulting in \( \ln |u| + C \).
- Step 4: Substitute back to \( x^2 + 1 \), finalizing as \( \frac{1}{2} \ln |x^2 + 1| + C \).
Trigonometric Substitution
Trigonometric substitution is another brilliant technique that leverages trigonometric identities to simplify integrals, particularly useful when dealing with expressions of the form \( \sqrt{a^2 + x^2} \), \( \sqrt{a^2 - x^2} \), or \( \sqrt{x^2 - a^2} \). For this exercise, the expression \( x^2 + 1 \) fits well with trigonometric substitution.
- Step 1: Let \( x = \tan \theta \), implying \( dx = \sec^2 \theta \, d\theta \) and hence \( x^2 + 1 = \sec^2 \theta \).
- Step 2: Substitute into the integral: \( \int \frac{\tan \theta \sec^2 \theta \, d\theta}{\sec^2 \theta} = \int \tan \theta \, d\theta \).
- Step 3: Integrate \( \tan \theta \, d\theta \) results in \( -\ln |\cos \theta| + C \).
- Step 4: With \( \theta = \tan^{-1}(x) \) and knowing \( \cos \theta = \frac{1}{\sqrt{x^2+1}} \), substitute back into the result.
Definite Integrals
Definite integrals are a powerful tool for calculating the area under a curve within a specified range. While our original exercise concerned indefinite integrals, understanding definite integrals is pivotal when the problem asks for specific bounds. When solving a definite integral, such as \( \int_a^b f(x) \, dx \), we use the antiderivative \( F(x) \) and apply:
- Evaluate \( F(b) - F(a) \) to determine the area under the curve from \( x = a \) to \( x = b \).
- Ensure functions are continuous on the interval \([a, b]\) for accuracy.
- For functions crossing the x-axis, consider the absolute value to find total area.
Antiderivatives
The concept of antiderivatives, or indefinite integrals, is central to integral calculus. In essence, finding an antiderivative means identifying a function whose derivative is equal to the given function. In other words, if \( F'(x) = f(x) \), then \( F(x) \) is an antiderivative of \( f(x) \).
- Antiderivatives involve adding a constant \( C \), as differentiation of a constant is zero, rendering it indistinguishable in derivatives.
- Gaining proficiency requires recognizing common derivatives and their corresponding antiderivatives.
- Techniques include reverse power rule, integration by parts, and trigonometric identities.
