Question

Use the method of partial fraction decomposition to perform the required integration. \(\int_{1}^{5} \frac{3 x+13}{x^{2}+4 x+3} d x\)

Short answer

The definite integral evaluates to \( \ln \left( \frac{243}{4} \right) \).

Step-by-step solution

Factor the denominator

First, we need to factor the denominator of the integrand. The denominator is a quadratic expression: \[ x^2 + 4x + 3 \]To factor it, we look for two numbers that multiply to 3 (the constant term) and add to 4 (the coefficient of the linear term):\[ x^2 + 4x + 3 = (x + 1)(x + 3) \]

Set up the partial fraction decomposition

Next, we'll express the integrand as a sum of partial fractions. Based on the factors from Step 1, the expression for the partial fractions is:\[ \frac{3x + 13}{x^2 + 4x + 3} = \frac{A}{x+1} + \frac{B}{x+3} \]where \( A \) and \( B \) are constants to be determined.

Solve for constants A and B

To find \( A \) and \( B \), rewrite the equation:\[ 3x + 13 = A(x+3) + B(x+1) \]Expand and simplify:\[ 3x + 13 = (A + B)x + (3A + B) \]Equating coefficients, we have:1) \( A + B = 3 \)2) \( 3A + B = 13 \)Solve this system of equations:From 1), \( B = 3 - A \).Substitute into 2): \( 3A + (3 - A) = 13 \)Simplify to find \( A \):\[ 3A - A = 10 \]\[ 2A = 10 \]\[ A = 5 \]Substitute \( A = 5 \) back into \( B = 3 - A \):\[ B = 3 - 5 = -2 \]So, \( A = 5 \) and \( B = -2 \).

Write the integrand as partial fractions

Replace the original integrand with its equivalent partial fraction decomposition using \( A = 5 \) and \( B = -2 \):\[ \frac{3x + 13}{x^2 + 4x + 3} = \frac{5}{x+1} - \frac{2}{x+3} \]

Integrate term by term

Now, integrate the expression term by term:\[ \int \left( \frac{5}{x+1} - \frac{2}{x+3} \right) dx = 5 \int \frac{1}{x+1} dx - 2 \int \frac{1}{x+3} dx \]Calculate each integral:1) \( \int \frac{1}{x+1} dx = \ln|x+1| + C \)2) \( \int \frac{1}{x+3} dx = \ln|x+3| + C \)So, the integral becomes:\[ 5 \ln|x+1| - 2 \ln|x+3| + C \]

Evaluate the definite integral

Compute the definite integral from \( x = 1 \) to \( x = 5 \):\[ \left[ 5 \ln|x+1| - 2 \ln|x+3| \right]_1^5 \]Calculate the expression at the upper and lower limits:\( x = 5: \)\[ 5 \ln|5+1| - 2 \ln|5+3| = 5 \ln 6 - 2 \ln 8 \]\( x = 1: \)\[ 5 \ln|1+1| - 2 \ln|1+3| = 5 \ln 2 - 2 \ln 4 \]Subtracting gives:\[ \left( 5 \ln 6 - 2 \ln 8 \right) - \left( 5 \ln 2 - 2 \ln 4 \right) \]Simplify using properties of logs:\[ 5(\ln 6 - \ln 2) - 2(\ln 8 - \ln 4) \]\[ = 5 \ln 3 - 2 \ln 2 \]This simplifies further to:\[ = \ln 243 - \ln 4 \]\[ = \ln \left( \frac{243}{4} \right) \]

Present the final answer

The value of the definite integral \( \int_{1}^{5} \frac{3x+13}{x^2+4x+3} dx \) is:\[ \ln \left( \frac{243}{4} \right) \]

Key concepts

Integral Calculus

Integral calculus is a cornerstone of mathematical analysis and deals with the accumulation of quantities. It's the process of finding integrals, which are essentially the reverse process of derivatives in calculus. The central idea lies in understanding how quantities add up and change over an interval to provide a total value or area under a curve.
The **indefinite integral** of a function is a family of functions whose derivative is the original function. Represented symbolically with an integral sign without bounds, it focuses on general solutions. On the other hand, the **definite integral** calculates the net area between the curve of a function and the x-axis over a specific interval, giving meaningful numerical results.
Understanding integral calculus involves several techniques, one of which is partial fraction decomposition, used especially for integrating rational functions. These methods allow for breaking down complex integrands into simpler terms that are easier to integrate. In essence, integral calculus provides powerful tools for solving problems involving areas and accumulations, crucial in physical sciences and engineering.

Definite Integrals

Definite integrals are an extension of the basic concept of integration, providing a precise way to calculate the area or accumulation of quantities over a defined interval. This interval is specified by its lower and upper bounds, which are integral limits. The result is a single number, representing a physical quantity like area, volume, or even total distance traveled.
When calculating a definite integral, you must perform the integration process using antiderivatives. The symbol for definite integration is a stylized S, followed by the function and bounds: \[ \int_{a}^{b} f(x) \, dx \] Here, \( a \) and \( b \) are the lower and upper limits, respectively. The process involves evaluating the antiderivative at these boundaries. The **Fundamental Theorem of Calculus** plays a crucial role, bridging the concept of integration and differentiation. It states that differentiation and integration are inverse processes. Essentially, to solve definite integrals, obtain the antiderivative and subtract its value at the lower limit from its value at the upper limit.
These integrals have wide-ranging applications from calculating physical quantities like area and volume to solving multidisciplinary problems in fields like economics, statistics, and engineering.

Logarithmic Functions

Logarithmic functions are the inverses of exponential functions and hold significant importance in calculus, especially when working with integrals that involve division by polynomials. Their basic form is \( \log_b(x) \), which answers the question: 'To what power must \( b \) be raised to yield \( x \)?'
In calculus, natural logarithms \( (\ln x) \), with base \( e \approx 2.718 \), are predominantly used. They simplify integration involving rational functions, particularly those leading to partial fraction decomposition scenarios.
When integrating simple rational functions like \( \frac{1}{x} \), the antiderivative is the natural logarithm function \( \ln|x| + C \), highlighting the connection between logarithms and integration.
This concept further extends to definite integrals over rational functions. In our example, logarithms appeared during integration after decomposing the function into partial fractions. This reveals the importance of understanding the properties of logarithms, such as \( \ln(ab) = \ln(a) + \ln(b) \) and \( \ln(\frac{a}{b}) = \ln(a) - \ln(b) \), as these properties help simplify expressions and solve integrals effectively. Through these techniques, logarithmic functions enable more straightforward solutions to integration problems and are indispensable in applied mathematics, science, and engineering contexts.