Chapter 7: Problem 42
Calculate the integrals.$$\int \frac{d z}{z^{2}+z}$$
Short Answer
Expert verified
The integral is \( \ln \left( \frac{|z|}{|z+1|} \right) + C \).
Step by step solution
01
Simplify the integrand
The first step in solving the integral is to simplify the integrand. The given integrand is \( \frac{1}{z^{2}+z} \). We can factor the denominator as \( z(z+1) \). Thus, the integral becomes: \[ \int \frac{1}{z(z+1)} \, dz \]
02
Use partial fraction decomposition
Since the denominator can be written in factored form, we can use partial fraction decomposition. We decompose \( \frac{1}{z(z+1)} \) into:\[ \frac{1}{z(z+1)} = \frac{A}{z} + \frac{B}{z+1} \]We need to find the constants \( A \) and \( B \) such that:\[ \frac{1}{z(z+1)} = \frac{A(z+1) + Bz}{z(z+1)} \]This implies:\[ 1 = A(z+1) + Bz \]Expanding and rearranging gives:\[ 1 = Az + A + Bz \] Combine like terms:\[ 1 = (A + B)z + A \]From this equation, we get two simultaneous equations:- \( A + B = 0 \)- \( A = 1 \)
03
Solve for constants A and B
From the equations derived in Step 2, substitute \( A = 1 \) into the first equation:\[ 1 + B = 0 \]Solving for \( B \), we get \( B = -1 \).Therefore, our partial fractions are:\[ \frac{1}{z(z+1)} = \frac{1}{z} - \frac{1}{z+1} \]
04
Integrate the partial fractions
Now, integrate each term separately:\[ \int \left( \frac{1}{z} - \frac{1}{z+1} \right) \, dz = \int \frac{1}{z} \, dz - \int \frac{1}{z+1} \, dz \]Each integral is a basic logarithm derivative:\[ \int \frac{1}{z} \, dz = \ln |z| + C_1 \]\[ \int \frac{1}{z+1} \, dz = \ln |z+1| + C_2 \]Combine these results:\[ \ln |z| - \ln |z+1| + C \]This simplifies to:\[ \ln \left( \frac{|z|}{|z+1|} \right) + C \]
05
Combine the expression
The processed integral with combined constants is: \[ \ln \left( \frac{|z|}{|z+1|} \right) + C \]where \( C \) is the constant of integration.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Partial Fraction Decomposition
Partial fraction decomposition is a technique in integral calculus for simplifying the integration of rational functions. A rational function is a ratio of two polynomials. When you encounter a rational function like \( \frac{1}{z^2 + z} \), start by factoring the denominator.
In our example, the denominator factors as \( z(z + 1) \). Partial fraction decomposition allows you to express this simplified function as a sum of simpler fractions. In this case:
With these simple fractions, the task of integrating the function is much more manageable.
In our example, the denominator factors as \( z(z + 1) \). Partial fraction decomposition allows you to express this simplified function as a sum of simpler fractions. In this case:
- We write \( \frac{1}{z(z+1)} \) as \( \frac{A}{z} + \frac{B}{z+1} \)
- The goal is to find constants \( A \) and \( B \) that satisfy the original equation.
With these simple fractions, the task of integrating the function is much more manageable.
Logarithmic Integration
Logarithmic integration is a common method used when integrating functions that result in logarithmic forms. A common scenario in which this arises is when the integrand involves fractions of the form \( \frac{1}{x} \). For example, integrating \( \frac{1}{z} \) results in \( \ln |z| + C \), where \( C \) is the constant of integration.
In our original problem, once we split \( \frac{1}{z(z+1)} \) into its partial fractions, it becomes easier to integrate:
In our original problem, once we split \( \frac{1}{z(z+1)} \) into its partial fractions, it becomes easier to integrate:
- For \( \int \frac{1}{z} \, dz \): The antiderivative is \( \ln |z| + C_1 \).
- For \( \int \frac{1}{z+1} \, dz \): The antiderivative is \( \ln |z+1| + C_2 \).
Rational Functions
Rational functions are a core topic in integral calculus and concern expressions that are ratios of polynomials. They can be constant, linear, quadratic, or higher-degree polynomials, over another polynomial. These kinds of functions can be challenging to integrate in their original form.
For instance, the function \( \frac{1}{z^2 + z} \) is a simple rational function. The key to integrating such functions is simplification, often through partial fraction decomposition as illustrated earlier.
For instance, the function \( \frac{1}{z^2 + z} \) is a simple rational function. The key to integrating such functions is simplification, often through partial fraction decomposition as illustrated earlier.
- Start with factoring the elements whenever possible.
- Decompose the fraction to split it into simpler components.