/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 42 Calculate the integrals.$$\int \... [FREE SOLUTION] | 91Ó°ÊÓ

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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:
  • 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.
By expanding and equating coefficients, you'll get a system of linear equations. Solving these equations gives us \( A = 1 \) and \( B = -1 \). Thus, \( \frac{1}{z(z+1)} = \frac{1}{z} - \frac{1}{z+1} \). This decomposition simplifies the integration process significantly.

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:
  • For \( \int \frac{1}{z} \, dz \): The antiderivative is \( \ln |z| + C_1 \).
  • For \( \int \frac{1}{z+1} \, dz \): The antiderivative is \( \ln |z+1| + C_2 \).
These logarithmic results occur because the derivative of the logarithmic function \( \ln |x| \) is \( \frac{1}{x} \). This process of integrating logarithmically is straightforward when dealing with integrands that are reducible to simple fractions.
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.

  • Start with factoring the elements whenever possible.
  • Decompose the fraction to split it into simpler components.
Once decomposed, the functions become sums or differences of simpler rational terms, like \( \frac{1}{z} \) and \( \frac{1}{z+1} \), which are easily integrable. Techniques like these reduce complex operations into simpler arithmetic, facilitating easier integration.

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