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Perform the indicated integrations. $$ \int \frac{x}{x^{2}+4} d x $$

Short Answer

Expert verified
\( \frac{1}{2} \ln(x^2 + 4) + C \)

Step by step solution

01

Identify the Integration Technique

This integral can be solved using the substitution method. We will identify a substitution that simplifies the integrand.
02

Choose a Suitable Substitution

Let \( u = x^2 + 4 \). Then, differentiate both sides with respect to \( x \) to find \( du \). We get \( du = 2x \, dx \). Therefore, \( x \, dx = \frac{1}{2} \, du \).
03

Transform the Integral

By substituting \( u = x^2 + 4 \) and \( x \, dx = \frac{1}{2} \, du \), the integral becomes: \[ \int \frac{x}{x^2+4} \, dx = \int \frac{1}{2} \frac{du}{u} \].
04

Integrate Using the Substitution

The integral \( \int \frac{1}{2} \frac{du}{u} \) simplifies to \( \frac{1}{2} \int \frac{du}{u} \), which is \( \frac{1}{2} \ln |u| + C \), where \( C \) is the constant of integration.
05

Substitute Back the Original Variable

Replace \( u \) with \( x^2 + 4 \) to express the integral in terms of \( x \): \( \frac{1}{2} \ln |x^2 + 4| + C \).
06

Simplify the Expression

The logarithmic absolute value does not affect the natural domain since \( x^2 + 4 > 0 \) for all real \( x \). Thus, the simplified final answer is \( \frac{1}{2} \ln(x^2 + 4) + C \).

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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 is a powerful tool in integral calculus used to simplify integrals, making them easier to solve. This method involves changing variables in the integral to transform it into a more straightforward form.
Here's how to effectively use the substitution method:
  • Identify a part of the integrand that can be substituted. This is usually an expression that, when replaced, simplifies the integrand considerably.
  • Assign a new variable to this expression. For instance, in our integral \( \int \frac{x}{x^2+4} \, dx \), we choose \( u = x^2 + 4 \).
  • Differentiate the new variable with respect to \( x \) to find \( du \). Here, \( du = 2x \, dx \), which implies \( x \, dx = \frac{1}{2} \, du \).
  • Replace the identified part and its differential in the integral with the new variable and its differential.
This approach simplifies the integral, allowing it to be solved more readily than in its original form. Once the substitution has simplified the integral, solve and revert back to the original variable.
Integration Techniques
Integration techniques are crucial in calculus to enable the solving of complex integrals. Several methods exist, with substitution being one among them.
When tackling an integral:
  • First, inspect whether the integrand hints at any recognizable transformations, like substitution or parts.
  • In substitution, you're simplifying structure by temporarily changing variables. For example, \( \int \frac{x}{x^2+4} \, dx \) appears complex, yet becomes simple with a suitable substitution.
  • Integration by parts or partial fraction decomposition may also be applicable for different types of integrals, enhancing the breadth of tools available.
Mastering these techniques requires practice since each integral may need a unique approach. Once comfortable with a range of techniques, integrating becomes more of a creative problem-solving task.
Logarithmic Integration
Logarithmic integration is a technique often encountered when integrating functions of the form \( \frac{1}{x} \) or expressions that transform to this structure through substitution.
Here’s how it works:
  • After performing a substitution like \( u = x^2 + 4 \), many integrals reduce to a form resembling \( \int \frac{du}{u} \).
  • This results in a natural logarithm: \( \, \ln|u| \) after integration.
  • It is critical to include the absolute value within the logarithm to account for negative values of \( u \). However, in our problem, \( x^2 + 4 \) is always positive, simplifying \( \ln|x^2 + 4| \) to \( \ln(x^2 + 4) \).
Understanding the process of logarithmic integration is vital since it frequently appears in calculus problems, especially when dealing with rational functions.

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