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Partial fraction decomposition is a method used in calculus to break down complex rational expressions into simpler fractions. This process is very useful when we want to integrate rational functions. As seen in the given problem, the task is to decompose \( \frac{2}{x^2 + x} \) into simpler terms. A rational function is expressed as the ratio of two polynomials. The denominator \( x^2 + x \) can be split into linear factors. So, we express \( \frac{2}{x^2 + x} \) as the sum of fractions with unknown constants, \( \frac{A}{x} + \frac{B}{x+1} \). To find the values of \( A \) and \( B \), we'll clear the fractions by multiplying through by the common denominator, \( x(x+1) \). This leads to an equation that we can solve for the constants by substituting convenient values for \( x \). In this problem, substituting \( x = 0 \) and \( x = -1 \) makes finding \( A \) and \( B \) straightforward, resulting in \( A = 2 \) and \( B = -2 \). By using this method, the original expression is decomposed into \( \frac{2}{x} - \frac{2}{x+1} \).

Integration is the process of finding the integral or antiderivative of a function. It's the reverse process of differentiation and is used to determine the area under a curve or solve differential equations. In the problem, once we have decomposed \( \frac{2}{x^2 + x} \) into partial fractions \( \frac{2}{x} - \frac{2}{x+1} \), the next step is to integrate these simpler fractions separately.For each fraction, we apply the standard integration rules. - The integral of \( \frac{1}{x} \) with respect to \( x \) is \( \ln|x| \).- Similarly, the integral of \( \frac{1}{x+1} \) is \( \ln|x+1| \). Thus, integrating the entire expression gives:\[p(x) = 2 \ln|x| - 2 \ln|x+1| + C\]where \( C \) is the constant of integration, representing the family of solutions to the integral.

Differential equations involve functions and their derivatives, expressing rates of change and modeling how quantities evolve over time or space. They are fundamental in fields such as physics, engineering, and economics. This problem involves a first-order differential equation, which contains the first derivative \( p'(x) \).The given differential equation \( p'(x) = \frac{2}{x^2 + x} \) is an example, and solving it involves finding a function \( p(x) \) such that the derivative equals the given expression. By integrating \( \frac{2}{x^2 + x} \) (after partial fraction decomposition), we seek a function whose rate of change matches the original differential equation.The solution to this equation captures the behavior of the system described by the differential equation. In practical terms, it tells us how \( p(x) \) changes with \( x \). The initial condition \( p(1) = 0 \) helps identify the particular solution within the family of solutions represented by the indefinite integral.

The constant of integration, \( C \), appears when taking the antiderivative of a function. It's an essential part of solving differential equations as it allows us to account for all possible antiderivatives. When integrating, we determine a general solution that includes \( C \), representing an infinite set of functions differing by a constant. Solving initial value problems (IVP) helps us find the particular solution out of this infinite set, by using given conditions. In our example, the initial value \( p(1) = 0 \) requires that we substitute \( x = 1 \) into the integrated equation \( 2 \ln|x| - 2 \ln|x+1| + C \). This lets us solve for \( C \) to satisfy the initial condition. Solving \( 0 = 2 \ln|1| - 2 \ln|2| + C \) gives \( C = 2 \ln 2 \). Including \( C \) aligns the integral with real-world phenomena or constraints by ensuring the solution is specific to the scenario described.