Problem 44 - Calculus Single Variable

Question

Calculate each of the definite integrals. $$ \int_{1}^{2} \frac{x^{2}-2 x-2}{x^{2}(x+1)} d x $$

Step-by-Step Solution

Answer

The integral evaluates to \( \ln \frac{3}{2} - 1 \).

Step 1: Simplify the Expression

First, we simplify the integral by performing partial fraction decomposition on the integrand \( \frac{x^{2}-2x-2}{x^{2}(x+1)} \). We express the fraction as:\[ \frac{x^{2}-2x-2}{x^{2}(x+1)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x+1} \]Multiply both sides by the denominator \(x^{2}(x+1)\) to clear the fractions:\[ x^2 - 2x - 2 = A x(x+1) + B(x+1) + C x^2 \]Then, equate coefficients to find \(A\), \(B\), and \(C\).

Step 2: Find the Coefficients

Expand and collect like terms:\[ x^2 - 2x - 2 = (A + C)x^2 + (A + B)x + B \]By comparing coefficients, we form the system of equations:1. \(A + C = 1\) (for \(x^2\))2. \(A + B = -2\) (for \(x\))3. \(B = -2\) (constant term)Using \(B = -2\) in equation 2, we find:\[A - 2 = -2 \quad \Rightarrow \quad A = 0\]Substitute \(A = 0\) in equation 1:\[C = 1\]Thus, \(A = 0\), \(B = -2\), \(C = 1\).

Step 3: Set Up the Integral with Partial Fractions

With \(A = 0\), \(B = -2\), and \(C = 1\), the partial fractions are:\[ \frac{x^{2}-2x-2}{x^{2}(x+1)} = \frac{-2}{x^2} + \frac{1}{x+1} \]This simplifies the integral:\[ \int_{1}^{2} \left(\frac{-2}{x^2} + \frac{1}{x+1}\right) \, dx \]

Step 4: Integrate the Partial Fractions

Integrate each term separately:- The integral of \(\frac{-2}{x^2}\) is: \[ \int \frac{-2}{x^2} \, dx = 2\cdot \frac{1}{x} = \frac{2}{x} \]- The integral of \(\frac{1}{x+1}\) is: \[ \int \frac{1}{x+1} \, dx = \ln|x+1| \]So the whole integral becomes:\[ \left[ \frac{2}{x} + \ln|x+1| \right]_{1}^{2} \]

Step 5: Evaluate the Definite Integral

Evaluate the antiderivative at the bounds:\[ \left( \frac{2}{2} + \ln|2+1| \right) - \left( \frac{2}{1} + \ln|1+1| \right) \]This simplifies to:\[ 1 + \ln 3 - 2 - \ln 2 \]Combining like terms:\[ -1 + \ln \frac{3}{2} \]

Step 6: Final Result

The value of the definite integral is:\[ \int_{1}^{2} \frac{x^{2}-2x-2}{x^{2}(x+1)} \, dx = \ln \frac{3}{2} - 1 \].

Key Concepts

Partial Fraction Decomposition

Partial fraction decomposition is a method used to break down complex rational expressions into simpler fractions. This technique is especially helpful in integral calculus when dealing with complex algebraic fractions as it simplifies the integration process.
When you are given a fraction in the form \( \frac{P(x)}{Q(x)} \) where \( P(x) \) and \( Q(x) \) are polynomials, and the degree of \( P(x) \) is less than the degree of \( Q(x) \), you can express it as a sum of simpler fractions.

• The denominator \( Q(x) \) is factored completely into linear and quadratic terms that aren't easily reducible.
• Each factor gives rise to a partial fraction with undetermined coefficients.
• We then solve for these coefficients to complete the decomposition.

In the integral \( \int_{1}^{2} \frac{x^{2}-2x-2}{x^{2}(x+1)} \, dx \), the decomposition was set as \( \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x+1} \), and solved using algebraic techniques.

Integral Calculus

Integral calculus focuses on finding the integral of functions, which is essentially the reverse process of differentiation. There are two main types of integrals:

• Indefinite integrals which do not have limits and find the general form of antiderivatives.
• Definite integrals involve limits and calculate the net area under a curve from one point to another.

In this exercise, we worked with a definite integral: \[ \int_{1}^{2} \left(\frac{-2}{x^2} + \frac{1}{x+1}\right) \, dx \] After decomposition and setting up the integral, it’s critical to integrate each term separately. Computing definite integrals involves evaluating the antiderivative at the upper and lower bounds to find the net area, using the Fundamental Theorem of Calculus.
For instance, integrating \( \frac{-2}{x^2} \), yields \( \frac{2}{x} \); while integrating \( \frac{1}{x+1} \) produces \( \ln|x+1| \). Evaluating these expressions over the interval \([1, 2]\) provides the final result.

Coefficient Comparison

Coefficient comparison is a fundamental strategy used to determine the unknown constants in partial fraction decomposition.
By expanding both sides of an equation containing the polynomial parts, you align terms of the same degree and compare their coefficients. This technique allows you to form a system of linear equations that can be solved to identify the unknown coefficients.

• Consider the expanded form: \( x^2 - 2x - 2 = (A + C)x^2 + (A + B)x + B \).
• The coefficients for each power of \( x \) on both sides are equated, creating a system of equations.
• Solving these equations using simple algebra helps find the values of \( A, B, \) and \( C \).

In this example, comparing coefficients resulted in equations like \( B = -2 \), \( A + B = -2 \), and \( A + C = 1 \). Solving these gave the final coefficients \( A = 0, B = -2, C = 1 \), which are used in the partial fraction expansion.