Problem 42 - Calculus Single Variable

Question

Calculate each of the definite integrals. $$ \int_{0}^{1} \frac{1}{(x+1)(2 x+1)} d x $$

Step-by-Step Solution

Answer

The integral evaluates to \( \ln \left( \frac{3}{2} \right) \).

Step 1: Decompose the fraction using partial fraction

To solve the integral \( \int_{0}^{1} \frac{1}{(x+1)(2x+1)} \, dx \), we first express the integrand as a sum of partial fractions. We have:\[\frac{1}{(x+1)(2x+1)} = \frac{A}{x+1} + \frac{B}{2x+1}\]Where \(A\) and \(B\) are constants that we need to determine.

Step 2: Set up the equation for partial fractions

Multiply through by the common denominator \((x+1)(2x+1)\) to clear the fractions:\[1 = A(2x+1) + B(x+1)\] Now distribute and expand:\[1 = (2A)x + A + Bx + B\] This simplifies to:\[1 = (2A + B)x + (A + B)\] Since this must hold for all \(x\), equate coefficients with the terms of the polynomial on the right-hand side to those on the left.

Step 3: Solve the system of linear equations

From the equation \[(2A + B)x + (A + B) = 1\]We equate coefficients:1. \(2A + B = 0\)2. \(A + B = 1\)Solve these equations simultaneously.From equation 1: \(B = -2A\), plug into equation 2:\[A - 2A = 1\]\[-A = 1\]\[A = -1\]Substitute \(A = -1\) back into \(B = -2A\):\[B = -2(-1) = 2\]

Step 4: Substitute back to the integrand

Substitute \( A = -1 \) and \( B = 2 \) into the partial fractions:\[\frac{1}{(x+1)(2x+1)} = \frac{-1}{x+1} + \frac{2}{2x+1}\] Now our integral becomes two simpler integrals:\[\int_{0}^{1} \left( \frac{-1}{x+1} + \frac{2}{2x+1} \right) \, dx\]

Step 5: Integrate each term separately

Integrate each part separately:1. \(\int_{0}^{1} \frac{-1}{x+1} \, dx = -\ln|x+1| \Big|_0^1 = -\ln 2 + \ln 1 = -\ln 2 \)2. \(\int_{0}^{1} \frac{2}{2x+1} \, dx\): Let \(u = 2x + 1\), so \(du/dx = 2\) or \(dx = du/2\). Change the bounds: When \(x = 0\), \(u = 1\); when \(x = 1\), \(u = 3\). The integral becomes \(\int_{1}^{3} \frac{2}{u} \, \frac{du}{2}\) = \( \int_{1}^{3} \frac{1}{u} \, du = \ln|u| \Big|_1^3 = \ln 3 - \ln 1 = \ln 3\)

Step 6: Combine the results

Combine the results from both integrals:\[-\ln 2 + \ln 3\]This simplifies to:\[\ln 3 - \ln 2 = \ln \left( \frac{3}{2} \right)\] Therefore, the answer is \( \ln \left( \frac{3}{2} \right) \).

Key Concepts

Partial Fraction Decomposition

Partial fraction decomposition is a method used to express a complex rational expression as the sum of simpler fractions. This technique is particularly useful in integral calculus for simplifying integrands before integration. To perform a partial fraction decomposition, you first need to identify the denominator of the fraction and set it up as a result of multiple simpler fractions.

For instance, to decompose \( \frac{1}{(x+1)(2x+1)} \), we assume it equals \( \frac{A}{x+1} + \frac{B}{2x+1} \). Here, \(A\) and \(B\) are constants to be determined. Next, multiply through by the common denominator to eliminate the fractions, resulting in an equation like \( 1 = A(2x+1) + B(x+1) \). Expand and distribute these terms to form a polynomial, then match the coefficients on both sides to set up a system of linear equations.

• Choose the correct form of partial fractions based on the factors of the denominator.
• Multiply through by the common denominator to setup a polynomial equation.
• Solve for the constants by equating like coefficients.

Once you have \(A\) and \(B\), substitute them back into the partial fractions to simplify the integrand, making it easier to integrate.

Integral Calculus

Integral calculus is a branch of mathematics focused on the process of integration, which is the reverse of differentiation. The main purpose of integral calculus is to find the total accumulated value, areas under curves, and antiderivatives of functions. In the context of definite integrals, the goal is to evaluate the integral within specific bounds, providing a numerical value.

To evaluate an integral, especially a definite one like \( \int_{0}^{1} \frac{1}{(x+1)(2x+1)} \, dx \), you might first decompose the integrand using partial fraction decomposition to simplify the expression, as shown in the previous step. Then, you integrate each simpler part independently using known integral formulas or techniques such as substitution.

• Break complex functions into simpler parts, if possible.
• Use definite integral properties to calculate the area or accumulated value.
• Apply substitution techniques when necessary to handle more complex expressions.

After integrating, apply the bounds to the antiderivative to calculate the definite integral. Finally, simplify the result.

System of Linear Equations

A system of linear equations is a collection of two or more equations with the same set of variables. When working with partial fraction decomposition, you often end up setting a system of equations to solve for unknown coefficients. This step is crucial for simplifying the integrand before continuing with integration.

For example, when decomposing \( \frac{1}{(x+1)(2x+1)} \), you get the equation \( 1 = A(2x+1) + B(x+1) \). Expanding this and equating coefficients provides a system of linear equations like\( 2A + B = 0 \) and \( A + B = 1 \). Solving these equations involves:

• Substituting one equation into another to eliminate variables.
• Simplifying the resulting equations to find the value of each constant.
• Substitute the solutions back into the original equations to verify correctness.

Solving this system provides the values for \(A\) and \(B\), which are plugged back into the partial fractions to continue the process of integration.