91Ó°ÊÓ

Partial fraction decomposition is a crucial technique in integral calculus. It simplifies the integration of rational functions, those defined as the ratio of two polynomials. The goal is to break a complex fraction into simpler pieces that are easier to integrate.

In our original exercise, we use partial fraction decomposition on \( \frac{2}{x(x^2+6)} \). We express this fraction in terms of simpler fractions. This approach allows us to consider the integral of each part separately. Here's how the process generally works:

• First, identify that the denominator can be factored. In this case, \( x(x^2 + 6) \) is already factored.
• Set up the decomposition expression. For a denominator like \( x(x^2 + 6) \), we choose \( \frac{A}{x} + \frac{Bx+C}{x^2+6} \) as the form.
• Multiply through by the common denominator to eliminate fractions and simplify.
• Match coefficients with the original expression to solve for \( A \), \( B \), and \( C \).

This decomposition breaks down our integral into terms of \( \frac{1}{3x} \) and \( -\frac{x}{3(x^2+6)} \), making the problem easier to handle.

The substitution method is another indispensable technique in calculus for handling integrals. It involves substituting a part of the integrand with a new variable to simplify the expression.

For instance, in the exercise, we use substitution for the term \( \frac{3}{x^2 + 6} \). We set \( u = x^2 + 6 \) and calculate the differential \( du = 2x \, dx \). This change transforms the integral into a related, but simpler form that is easier to solve:

• Choose a substitution that simplifies the integrand. Here, \( u = x^2 + 6 \) is sensible because it represents a composite function within the integral.
• Determine \( du \) by differentiating the substitution equation with respect to \( x \).
• Replace all \( x \)-related terms with \( u \). This often includes adjusting the \( dx \) as \( x \) terms usually need to be expressed in terms of \( u \) as well, as done here \( dx = \frac{du}{2x} \).

Once the substitution is complete, the integral can be rewritten as \( \int \frac{3}{u} \, du \), which is a simple logarithmic integral form, yielding \( 3\ln|u| \), and ultimately, \( 3\ln|x^2 + 6| \) after reverting back to \( x \).

Indefinite integrals represent the collection of all antiderivatives of a function. They are written as \( \int f(x) \, dx \) and include an arbitrary constant \( C \).

In this exercise, once we've simplified the integrand using techniques like partial fraction decomposition and substitution, we proceed to evaluate indefinite integrals for each part. Here's a look at each step:

• For \( \frac{3}{x^2 + 6} \), using substitution, we find the antiderivative to be \( 3\ln|x^2 + 6| \).
• For \( \frac{1}{3x} \), the integral is a straightforward case of finding the natural logarithm, resulting in \( \frac{1}{3}\ln|x| \).
• Lastly, for \( -\frac{x}{3(x^2 + 6)} \), after substitution, the integral simplifies to \( -\frac{1}{6}\ln|x^2 + 6| \).

These indefinite integrals showcase how we can reverse the process of differentiation. Every indefinite integral, therefore, represents a family of functions, differing by a constant \( C \). Combining these results provides the complete solution, ensuring the constant represents the general nature of the antiderivative.