Problem 40 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 1–40, use the method of partial fraction decomposition to perform the required integration. $$ \int_{1}^{5} \frac{3 x+13}{x^{2}+4 x+3} d x $$

Step-by-Step Solution

Answer

The integral evaluates to \(\ln18\).

Step 1: Factor the Denominator

The first step in partial fraction decomposition is factoring the denominator. The denominator is a quadratic expression: \(x^2 + 4x + 3\). We can factor it into \((x + 1)(x + 3)\).

Step 2: Setup Partial Fraction Decomposition

We express the fraction \( \frac{3x + 13}{(x+1)(x+3)} \) as the sum of two partial fractions. This means: \( \frac{A}{x+1} + \frac{B}{x+3} \). We need to solve for constants \(A\) and \(B\).

Step 3: Solve for A and B

To find \(A\) and \(B\), rewrite:\[ \frac{3x + 13}{(x+1)(x+3)} = \frac{A(x+3) + B(x+1)}{(x+1)(x+3)} = \frac{(A+B)x + (3A + B)}{(x+1)(x+3)} \].Equate: \(3x + 13 = (A+B)x + (3A + B)\).Matching coefficients gives: \(A + B = 3\) and \(3A + B = 13\). Solve these equations to get \(A = 2\) and \(B = 1\).

Step 4: Integrate Each Partial Fraction

With \(A = 2\) and \(B = 1\), the integral becomes: \( \int_{1}^{5} \frac{2}{x+1} dx + \int_{1}^{5} \frac{1}{x+3} dx \).Integrate each term separately:\[ \int \frac{2}{x+1} \, dx = 2\ln|x+1| + C \].\[ \int \frac{1}{x+3} \, dx = \ln|x+3| + C \].

Step 5: Evaluate Definite Integrals

Now evaluate from 1 to 5:For \(\int_{1}^{5} \frac{2}{x+1} \, dx \):\[ 2\ln|5+1| - 2\ln|1+1| = 2\ln6 - 2\ln2 = 2(\ln6 - \ln2) \].For \(\int_{1}^{5} \frac{1}{x+3} \, dx \):\[ \ln|5+3| - \ln|1+3| = \ln8 - \ln4 = \ln(8/4) = \ln2 \].

Step 6: Combine Results and Simplify

Combine the evaluated integrals:\[ 2(\ln6 - \ln2) + \ln2 = 2\ln3 + \ln2 = \ln(3^2) + \ln2 = \ln(9 \times 2) = \ln18 \]. So the answer is \( \ln18 \).

Key Concepts

Integration Techniques

Integration is more than just a mathematical process; it's a powerful tool that helps us understand and calculate areas under curves, among many other things. In this exercise, we used a specific integration technique known as "partial fraction decomposition." This method is especially useful when dealing with rational functions, where we can rewrite complex fractions as a sum of simpler fractions.
To successfully apply partial fraction decomposition, we first factor the denominator, if possible. This exercise involves the fraction \( \frac{3x+13}{x^2+4x+3} \), where the denominator factors neatly into \((x+1)(x+3)\). After factoring, we express our original fraction as a series of partial fractions, such as \( \frac{A}{x+1} + \frac{B}{x+3} \).
This decomposition is not just an algebraic trick; it makes integration simpler. By breaking down a complicated fraction, each part becomes an easier target for specific integration techniques associated with logarithmic functions. After setting up the partial fractions, we solve for constants like \(A\) and \(B\) using algebraic equations, allowing us to integrate each part separately.

Definite Integrals

This problem specifically targets finding the definite integral, which means we aim to calculate the exact area under the curve between specific bounds—in this case, between \(x=1\) and \(x=5\).
The definite integral \( \int_{a}^{b} f(x) \, dx \) gives the net area between the curve \( f(x) \) and the x-axis from \( x=a \) to \( x=b \). In our exercise, after decomposing the fraction, we are left with two simpler integrals \( \int \frac{2}{x+1} \, dx \) and \( \int \frac{1}{x+3} \, dx \).
Once we've integrated these individual fractions, we need to evaluate them at the upper and lower limits to get the net area. This part of the process involves substituting back the limits to get precise numerical results, as we calculated with \( 2(\ln6 - \ln2) + \ln2 \), which simplifies to \( \ln18 \). Thus, definite integrals give us exact values, which are incredibly useful in fields ranging from physics to economics.

Calculus Problems

Calculus problems vary widely in terms of complexity, but the underlying goal remains the same: understanding changes and areas. Problems like the one described involve integrating a rational function, offering a hands-on way to practice integration techniques like partial fraction decomposition.
Such problems help solidify understanding of the properties and rules governing calculus. Through incremental steps—like factoring, solving for constants, integrating part by part, and evaluating the definite integral—we gain insights into the behavior of functions.
Approaching these problems methodically is key. By breaking down each step and clearly understanding the rationale behind actions such as choosing a specific technique or recognizing when to apply a logarithmic function, students develop robust problem-solving skills. This disciplined approach not only helps in handling specific calculus problems but also builds a strong foundation for tackling more advanced topics in mathematics and applied sciences.