Question

Use the method of partial fractions to evaluate the following integrals. \(\int \frac{2}{(x+2)^{2}(2-x)} d x\)

Short answer

The integral evaluates to \(\frac{1}{2} \ln|x+2| + C.\)

Step-by-step solution

Set Up Partial Fractions

The first step is to express the function \(\frac{2}{(x+2)^{2}(2-x)}\) in terms of partial fractions. For this, assume the form \(\frac{A}{x+2} + \frac{B}{(x+2)^2} + \frac{C}{2-x}\). Multiply through by the common denominator \((x+2)^{2}(2-x)\) to get:\[2 = A(x+2)(2-x) + B(2-x) + C(x+2)^2.\]

Expand and Simplify

Expand the right hand side:\[A(x+2)(2-x) = A(x^2 + 2x - x^2 - 2x + 4) = A(4),\]\[B(2-x) = B(2-x),\]\[C(x+2)^2 = C(x^2 + 4x + 4).\]Combine like terms:\[2 = A(4) + B(2-x) + C(x^2 + 4x + 4).\] This simplifies to:\[2 = Cx^2 + (4C - B)x + (4A + 4C + 2B).\]

Solve for Coefficients

Since there is no \(x^2\) term or \(x\) term on the left-hand side, the coefficients must satisfy:- \(C = 0\),- \(4C - B = 0\),- \(4A + 4C + 2B = 2\).Solve these equations. From \(C = 0\), substituting into the second equation gives \(-B = 0\) or \(B = 0\). Substitute \(B = 0\) and \(C = 0\) into the third equation:\[4A = 2 \Rightarrow A = \frac{1}{2}.\]

Write Partial Fraction Decomposition

Substitute \(A = \frac{1}{2}, B = 0,\) and \(C = 0\) back into the partial fraction form:\[\frac{2}{(x+2)^{2}(2-x)} = \frac{1/2}{x+2}.\] Thus, we have the decomposition:\[\frac{2}{(x+2)^2(2-x)} = \frac{1/2}{x+2}.\]

Integrate Each Term

Now, integrate the decomposed terms:\[\int \frac{1/2}{x+2} \, dx = \frac{1}{2} \int \frac{1}{x+2} \, dx = \frac{1}{2} \ln|x+2| + C.\]

Final Solution

The integral \(\int \frac{2}{(x+2)^{2}(2-x)} \, dx\) is \(\frac{1}{2} \ln|x+2| + C.\)

Key concepts

Integration

Integration is a fundamental tool in calculus used to find areas, volumes, central points, and many useful things. The core idea is to reverse the process of differentiation. In simpler terms, integration helps us determine the total accumulation of quantities. In the context of this problem, we're looking to compute the integral of a rational function. This type of function features a polynomial in its numerator and a polynomial in its denominator.

Because direct integration of complex rational functions can be challenging, calculus provides several strategies to simplify it. This is where the technique of using partial fractions becomes invaluable, transforming tricky integrals into manageable ones. As we’ll see, once the rational function is decomposed, integrating becomes a matter of handling simpler, separate terms. This makes the work easier and often leads directly to known results.

Rational Functions

Rational functions are fractions involving polynomials. Collectively, they form an essential part of calculus. In this exercise, our rational function is given by the integrand \(\frac{2}{(x+2)^2(2-x)}\). Simplifying this direct form into partial fractions is the key to finding the integral.

To break it down, understand the denominator as a product of simpler polynomial factors. The goal is to express the rational function as a sum of simpler fractions which are easier to integrate. Here, each factor from the denominator was assigned a potential numerator in the form of constants \(A\), \(B\), and \(C\). Such a transformation is guided systematically by algebra, ensuring all rational parts are covered.

The process of matching coefficients lets us resolve the constants, making sure the original equation holds true. It's like solving a puzzle where each piece must fit perfectly without changing the overall picture of the equation.

Calculus Techniques

Calculus is equipped with many techniques to make solving problems more efficient. Partial fraction decomposition is a prime example when tackling integrals of rational functions. By re-expressing an otherwise daunting expression into simpler components, calculus techniques break a problem down to digestible bits.

In this problem, we initially needed to set up the partial fractions by assuming possible numerators, translated through a lens of algebraic manipulation. This was followed by multiplying the terms to equate with the original rational expression.

• The expansion of terms was strategically used for identification.
• Matching coefficients served our discovery to isolate variables.
• Finally, solving for constants gave us the structure needed for straightforward integration.

Such techniques are beautifully intricate, a manifestation of how calculus turns complex ideas into solvable formats.

Definite Integrals

Definite integrals provide the total accumulative value between a start and end point along a function. However, for indefinite integrals, like the one in this problem, we are interested in finding a general form with a constant of integration. This means while we're technically not evaluating a specific range, understanding this principle is essential.

Once our rational function was decomposed into an easier form, the next task was to perform integration on each term separately. Here, the integral of \(\frac{1/2}{x+2}\) became straightforward due to its simpler nature.

In calculus, working with integrals often culminates in evaluating the antiderivative and adding a constant \(C\), which accounts for any number lacking context from previous derivations. It’s this flexibility and broad application which makes mastering integrals a cornerstone of calculus.