Problem 14 - Calculus for Biology and Medicine

Question

In Problems , use partial-fraction decomposition to evaluate the integrals. $$ \int \frac{1}{x(2 x+1)} d x $$

Step-by-Step Solution

Answer

\( \int \frac{1}{x(2x+1)} dx = \ln \left| \frac{x}{2x+1} \right| + C \).

Step 1: Identify the Type of Fraction

The given integrand is \( \frac{1}{x(2x+1)} \). Since the degree of the denominator is higher than the degree of the numerator, we can proceed with partial fraction decomposition.

Step 2: Set Up Partial Fractions

Express \( \frac{1}{x(2x+1)} \) as a sum of partial fractions: \( \frac{1}{x(2x+1)} = \frac{A}{x} + \frac{B}{2x+1} \).

Step 3: Solve for Coefficients

Multiply through by the common denominator, \( x(2x+1) \), to clear the fractions: \[ 1 = A(2x+1) + Bx \]. Expand and combine terms to get \[ 1 = (2A + B)x + A \].

Step 4: Equate and Solve the Coefficients

We have two equations by comparing coefficients: 1. \( 2A + B = 0 \) 2. \( A = 1 \)From equation 2, solve \( A = 1 \).Substitute \( A = 1 \) into equation 1: \( 2(1) + B = 0 \), so \( B = -2 \).

Step 5: Rewrite the Integral with Partial Fractions

Substitute \( A = 1 \) and \( B = -2 \) back into the partial fractions: \( \frac{1}{x(2x+1)} = \frac{1}{x} - \frac{2}{2x+1} \).Rewrite the integral as: \[ \int \left( \frac{1}{x} - \frac{2}{2x+1} \right) dx \].

Step 6: Integrate Each Term Separately

Integrate each term:- \( \int \frac{1}{x} dx = \ln |x| + C_1 \)- \( \int \frac{-2}{2x+1} dx = - \ln |2x+1| + C_2 \) (After applying substitution \( u = 2x+1 \), \( du = 2 dx \)).

Step 7: Combine Results and Simplify

Combine the results from Step 6:\[ \int \frac{1}{x(2x+1)} dx = \ln |x| - \ln |2x+1| + C \],where \( C = C_1 + C_2 \). Simplified as:\[ \ln \left| \frac{x}{2x+1} \right| + C \].

Key Concepts

Integral Calculus

Integral calculus is a branch of calculus focused on accumulating quantities, like areas under curves or the total growth represented by a rate of change. It's the process of finding integrals, which are essential to problems involving totality, such as problems encountered in physics, engineering, and economics.
When dealing with integrals, we often encounter fractions where the power of the denominator is higher than that of the numerator. This requires us to express the function in a simpler form known as partial fraction decomposition. This technique helps break down complex fractions into simpler terms, allowing us to integrate term-by-term more effectively.
Mastering integral calculus is crucial as it allows for the transition from infinitesimal changes to accumulated outcomes. In essence, it's all about understanding the impact of small parts on the whole function.

Solving Coefficients

The process of solving coefficients is a key part of partial fraction decomposition. This involves figuring out the values of constants that balance an equation once fractions are decomposed. In our exercise, we started with the expression:

• \(\frac{1}{x(2x+1)} = \frac{A}{x} + \frac{B}{2x+1} \)

We then multiplied through by the common denominator \(x(2x+1)\) to eliminate all fractions:
- \(1 = A(2x+1) + Bx\)
After distributing and aligning terms by matching coefficients, we ended up with two high school algebraic equations to solve:

• \(2A + B = 0\)
• \(A = 1\)

These equations revealed that \(A = 1\) and by substitution, \(B = -2\). Solving coefficients is foundational because it transforms the abstract decomposition into a directly usable form for integration.

Logarithmic Integration

Logarithmic integration is a technique used when the integral of a function results in a natural logarithm. After partial fraction decomposition, our integral becomes:
f- \( \int \left( \frac{1}{x} - \frac{2}{2x+1} \right) dx \) Each part of the separated integrand can now be integrated individually.
1. The integral of \( \frac{1}{x} \) is straightforward, resulting in \( \ln |x| + C_1 \). This follows from the fundamental rule of logarithmic differentiation.
2. For \( \frac{-2}{2x+1} \), a substitution method can simplify the process. Let \( u = 2x + 1 \), which implies \( du = 2 dx \). This substitution changes the integral into a standard logarithmic form: \(-2 \int \frac{1}{u} du = - \ln |2x+1| + C_2 \).
By combining these two results, we simplify the expression to log form: - \( \ln \left| \frac{x}{2x+1} \right| + C \) The constant \( C \) includes both constants from the separate integrations. Logarithmic integration not only adds a valuable tool to a mathematician's toolkit but also demonstrates the harmony between algebraic manipulation and integral solutions.