Chapter 7: Problem 2
Evaluate the given integral. $$\int \frac{x}{x^{2}+9} d x$$
Short Answer
Expert verified
The solution is \( \frac{1}{2} \ln (x^2 + 9) + C \).
Step by step solution
01
Identify the Integral Type
The given integral is \( \int \frac{x}{x^{2}+9} \, dx \). This is suitable for a substitution method because the denominator is a quadratic expression, and the numerator is its derivative.
02
Choose a Substitution
Let \( u = x^2 + 9 \). Then the derivative of \( u \) with respect to \( x \) is \( du = 2x \, dx \). We can rewrite this as \( \frac{1}{2} du = x \, dx \), which matches the structure of the integrand.
03
Rewrite the Integral
Substitute \( u \) and \( du \) into the integral: \( \int \frac{x}{x^2 + 9} \, dx = \int \frac{1}{2} \frac{1}{u} \, du \). This simplifies to \( \frac{1}{2} \int \frac{1}{u} \, du \).
04
Integrate
Integrate \( \frac{1}{u} \, du \) to obtain \( \ln |u| + C \). Therefore, the integral becomes \( \frac{1}{2} \ln |u| + C \).
05
Back-Substitute for Original Variable
Replace \( u \) by \( x^2 + 9 \) in the result of the integration from Step 4. Hence the integral becomes \( \frac{1}{2} \ln |x^2 + 9| + C \).
06
Simplify the Expression
Since \( x^2 + 9 \) is always positive for all real \( x \), there is no need for absolute values. Thus, the result is \( \frac{1}{2} \ln (x^2 + 9) + C \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Substitution Method
The substitution method in calculus is a powerful technique used to simplify integrals, particularly when dealing with complex expressions. It involves substituting part of the integrand with a new variable, which can make the integration process more straightforward. In our original problem, we faced the integral \( \int \frac{x}{x^2+9} \, dx \). The substitution method was apt because the numerator, \( x \), is similar to the derivative of the denominator, \( x^2 + 9 \). By letting \( u = x^2 + 9 \), we transformed the integral into a simpler form. This technique is particularly useful when:
- The integrand contains a composite function, and the inner function's derivative is present elsewhere in the integrand.
- There is a rational function where the substitution simplifies the integration.
- The integrand can be converted into a known integral form with the substitution.
Definite Integral
A definite integral computes the net area under a curve within a specific interval. It has a more precise application than an indefinite integral since it evaluates the function over a set range. In our exercise, though we were dealing with an indefinite integral, understanding definite integrals helps in recognizing the broader purpose of integration.
- A definite integral is denoted by \( \int_a^b f(x) \, dx \), where \( a \) and \( b \) are the lower and upper limits of integration.
- It provides a numerical value, unlike the indefinite integral, which yields a function plus a constant \( C \).
- Applying the Fundamental Theorem of Calculus is often necessary, linking integration with differentiation.
- Definite integrals can compute quantities such as areas, volumes, and even averages, making them incredibly beneficial in both theoretical and applied contexts.
Logarithmic Integration
Logarithmic integration is a technique that comes into play when integrating expressions resulting in a logarithmic function. It typically emerges when dealing with integrals like \( \int \frac{1}{u} \, du \), leading to results involving natural logarithms. In the case of our exercise, after substituting, we arrived at the integral \( \frac{1}{2} \int \frac{1}{u} \, du \). This simplifies to \( \frac{1}{2} \ln |u| + C \). Here's why logarithmic integration is helpful:
- It simplifies integrals where the function is in the form of \( \frac{1}{x} \), resulting directly in a logarithmic function when integrated.
- The natural log function \( \ln(x) \) is an antiderivative of \( \frac{1}{x} \), making it an elegant solution for these types of integrals.
- Logarithmic integration often emerges in problems involving growth, decay, and certain economic models.