Problem 79 - Calculus for Biology and Medicine

Question

Evaluate each definite integral. $$ \int_{3}^{5} \frac{x-1}{(x+1)(x+2)} d x $$

Step-by-Step Solution

Answer

The evaluated integral is \( \ln\left(\frac{343}{100}\right) \).

Step 1: Decompose Into Partial Fractions

First, recognize that \( \frac{x-1}{(x+1)(x+2)} \) needs to be expressed as a sum of simpler fractions for easier integration. Use partial fraction decomposition to express the integral as \( \frac{x-1}{(x+1)(x+2)} = \frac{A}{x+1} + \frac{B}{x+2} \). To find \(A\) and \(B\), equate the numerators:\[ x-1 = A(x+2) + B(x+1) \].Simplifying and equating coefficients will let us solve for the constants \(A\) and \(B\).

Step 2: Solve for Coefficients

Expand the equation from Step 1:\[ x-1 = Ax + 2A + Bx + B \]Grouping like terms gives:\[ x - 1 = (A + B)x + (2A + B). \]Thus, we have the system of equations:1. \( A + B = 1 \) (coefficient of \(x\))2. \( 2A + B = -1 \) (constant term).Solving, substitute \( B = 1 - A \) into the second equation:\[ 2A + (1 - A) = -1 \Rightarrow A + 1 = -1 \Rightarrow A = -2 \].Then, substitute \( A = -2 \) into \( B = 1 - A \):\[ B = 1 - (-2) = 3 \].Therefore, \( A = -2 \) and \( B = 3 \).

Step 3: Rewrite the Integral with Partial Fractions

Substitute \( A \) and \( B \) back into the partial fraction decomposition:\[ \frac{x-1}{(x+1)(x+2)} = \frac{-2}{x+1} + \frac{3}{x+2} \].Thus, the integral becomes:\[ \int_{3}^{5} \left( \frac{-2}{x+1} + \frac{3}{x+2} \right) \, dx \].

Step 4: Integrate Each Term Separately

Now, integrate each term separately:\[ \int_{3}^{5} \frac{-2}{x+1} \, dx = -2 \left[ \ln|x+1| \right]_{3}^{5} \]and\[ \int_{3}^{5} \frac{3}{x+2} \, dx = 3 \left[ \ln|x+2| \right]_{3}^{5} \].

Step 5: Evaluate the Definite Integrals

Substitute the limits of integration into each of the indefinite integrals calculated in Step 4:1. For \( \int_{3}^{5} \frac{-2}{x+1} \, dx \): \[ -2(\ln|5+1| - \ln|3+1|) = -2(\ln 6 - \ln 4) \] \[ = -2 \ln\left(\frac{6}{4}\right) = -2 \ln\left(\frac{3}{2}\right) \].2. For \( \int_{3}^{5} \frac{3}{x+2} \, dx \): \[ 3(\ln|5+2| - \ln|3+2|) = 3(\ln 7 - \ln 5) \] \[ = 3 \ln\left(\frac{7}{5}\right) \].

Step 6: Combine the Results

Combine the results from Step 5:\[ -2 \ln\left(\frac{3}{2}\right) + 3 \ln\left(\frac{7}{5}\right) \].Using logarithmic properties, we can combine:\[ = \ln \left( \left(\frac{7/5}{3/2}\right)^3 \right) = \ln\left(\frac{14}{10}\right)^3 = \ln\left(\frac{343}{100}\right) \].

Step 7: Final Step: Simplify and State the Answer

Simplify the expression:\( \ln\left(\frac{343}{100}\right) \).This is the result for the evaluated definite integral.

Key Concepts

Partial Fraction Decomposition

Partial fraction decomposition is a technique used to simplify complex fractions, making integration more manageable. In this method, we decompose a fraction into a sum of simpler fractions. This is particularly useful when dealing with rational functions, which are ratios of polynomials.

For the integral \( \int \frac{x-1}{(x+1)(x+2)} \ dx \), we can apply partial fraction decomposition to express \( \frac{x-1}{(x+1)(x+2)} \) as a sum \( \frac{A}{x+1} + \frac{B}{x+2} \).

• First, set up an equation based on the numerator: \( x-1 = A(x+2) + B(x+1) \).
• Next, expand and collect like terms, resulting in a system of equations to solve for \( A \) and \( B \).
• Solving this gives specific values for \( A \) and \( B \), which allows for rewriting the integral in terms of these simpler fractions.

This decomposed form simplifies the integration process as each term can now be integrated separately.

Integration Techniques

Integration is a fundamental concept in calculus used to find areas under curves and accumulation functions. Various techniques exist, each suited to different types of functions. Knowing which technique to apply is crucial.

After decomposing into partial fractions, \( \int \left( \frac{-2}{x+1} + \frac{3}{x+2} \right) dx \) involves basic integration of simple fractions:

• Each term is integrated separately, which is manageable due to the simpler form of the fractions.
• The antiderivative of \( \frac{1}{x+a} \) is \( \ln|x+a| \), a basic rule of integration.
• Constants, such as \(-2\) and \(3\), are factored out of the integrals, streamlining the process.

This results in two straightforward integrals that can be evaluated over the given limits to find the definite integral value.

Logarithmic Integration

Logarithmic integration often comes into play when dealing with integrals involving \( \frac{1}{x} \) or similar terms, resulting in a natural logarithm function as the antiderivative.

For our definite integral, integrating terms like \( \frac{-2}{x+1} \) and \( \frac{3}{x+2} \) leads to:

• \( \int \frac{-2}{x+1} \ dx = -2\ln|x+1| \)
• \( \int \frac{3}{x+2} \ dx = 3\ln|x+2| \)

These results rely on the rule that the integral of \( \frac{1}{x} \) is \( \ln|x| \).

Once the antiderivatives are determined, apply the boundaries from the definite integral, substituting and evaluating \( x \) values at both ends. The properties of logarithms can combine results, further simplifying the expression:
- Use subtraction within the logarithms: \( \ln(a) - \ln(b) = \ln\left(\frac{a}{b}\right) \)
This technique not only offers elegance but assists in evaluating the definite integral with precision.