91Ó°ÊÓ

Starting with an algebraic expression like \( \frac{7x^2 + 4x + 6}{(x+1)(x^2+2)} \), the goal of partial fraction decomposition is to break it down into simpler fractions. This makes the process of integration much more manageable. Partial fraction decomposition is a method used to express a complicated rational function as a sum of simpler fractions or partial fractions.
To begin, assume a decomposition of the integrand in the form \( \frac{A}{x+1} + \frac{Bx + C}{x^2+2} \). These are fractions with simpler denominators than the original function.

• Select the appropriate form based on the factors in the denominator.
• For distinct linear factors like \((x+1)\), use \( \frac{A}{x+1} \).
• For irreducible quadratic factors like \((x^2+2)\), use \( \frac{Bx + C}{x^2+2} \).

This step simplifies the integral into parts that can be solved more easily using basic integration techniques.

Once the partial fraction decomposition is assumed, the next step is to determine the constants \(A\), \(B\), and \(C\). This is done by setting up and solving a system of equations derived from equating coefficients.
Start by equating the expanded form of the sum of partial fractions to the original numerator:

• \(A(x^2 + 2) + (Bx + C)(x+1) = 7x^2 + 4x + 6\)
• Expand to get \( (A+B)x^2 + (B+C)x + (2A+C) = 7x^2 + 4x + 6 \)

By matching coefficients for each power of \(x\), we derive:

• \(A + B = 7\)
• \(B + C = 4\)
• \(2A + C = 6\)

These equations are solved systematically to find \(A = 3\), \(B = 4\), and \(C = 0\), providing the specific values needed for further integration efforts.

The substitution method, often called \(u\)-substitution, is a powerful technique in calculus for simplifying the integration process. This method involves replacing a part of the integrand with a new variable to simplify the integration.
For example, when dealing with the integral \( \int \frac{4x}{x^2+2} \, dx \), strategize by setting \( u = x^2 + 2 \). This transforms the integral, as the derivative \( du = 2x \, dx \) allows for a change of variables:

• Express the variable of integration in terms of \(u\).
• \( dx = \frac{du}{2x} \) lets you substitute back to \( u \).
• Integrate with respect to \( u \) after substituting.

This is an effective method because it simplifies otherwise complex integrals into more manageable forms, making evaluation straightforward.

The natural logarithm, \( \ln \), frequently appears in calculus, particularly in integration. When integrating terms like \( \frac{1}{x+1} \), the result often involves the natural logarithm.
In our case, consider the integral \( \int \frac{3}{x+1} \, dx \):

• On integrating, this yields \( 3\ln|x+1| \), with the absolute value ensuring the domain is correctly handled.

Next, evaluate the integral from 0 to 1:

• \( 3[\ln(2) - \ln(1)] = 3\ln(2) \).

Similarly, after substitution, integrals like \( \int \frac{4x}{x^2+2} \, dx \) also resolve into logarithmic expressions. After substitution, reaching the form \( 2\ln(\frac{3}{2}) \) shows the power of logs in simplifying calculations.
Understanding these results is crucial, as the properties of logarithms (like \( \log(a) + \log(b) = \log(ab) \)) are immensely useful in combining terms and simplifying solutions.