Question

Use substitution to convert the integrals to integrals of rational functions. Then use partial fractions to evaluate the integrals. \(\int \frac{1-\sqrt{x}}{1+\sqrt{x}} d x\)

Short answer

The integral is evaluated as \( 2\sqrt{x} - 2\ln|1+\sqrt{x}| - x \).

Step-by-step solution

Substitute to simplify the expression

To simplify the integral, use substitution. Let \( u = \sqrt{x} \). This means \( x = u^2 \) and \( \frac{dx}{du} = 2u \) so \( dx = 2u \, du \). Substitute these into the integral.

Simplify the integral with substitution

With the substitution, the integral \( \int \frac{1 - \sqrt{x}}{1 + \sqrt{x}} \, dx \) becomes \( \int \frac{1 - u}{1 + u} \cdot 2u \, du \), which simplifies to \( 2 \int \frac{u(1 - u)}{1 + u} \, du \).

Expand the integrand

Expand the integrand to prepare it for partial fraction decomposition: \( 2 \int \frac{u - u^2}{1 + u} \, du \). Divide each part by \(1+u\), resulting in \( 2 \int \left( u - u^2 \right) \frac{1}{1+u} \, du \).

Rewrite in a form suitable for partial fractions

Separate the expression as \( 2 \int \left( \frac{u}{1+u} - u^2 \frac{1}{1+u} \right) \, du \), leading to \( 2 \int \left( \frac{u}{1+u} \right) \, du - 2 \int \left( u \right) \, du \).

Solve first integral by partial fractions

The first integral, \( \int \frac{u}{1+u} \, du \), can be simplified directly to \( \int \left( 1 - \frac{1}{1+u} \right) \, du \) because \( \frac{u}{1+u} = 1 - \frac{1}{1+u} \). Solve this to get \( u - \ln|1+u| \).

Solve the second integral

The second integral is \( \int u \, du \), which is straightforward and equates to \( \frac{u^2}{2} \).

Combine the solutions and back-substitute

Combine the results from Steps 5 and 6, giving \( 2 \left( u - \ln|1+u| - \frac{u^2}{2} \right) \). Back-substitute \( u = \sqrt{x} \) to get \( 2 \left( \sqrt{x} - \ln|1+\sqrt{x}| - \frac{x}{2} \right) \).

Simplify the expression

Simplify to yield the final result: \( 2\sqrt{x} - 2\ln|1+\sqrt{x}| - x \).

Key concepts

Substitution Method

The substitution method is a key technique in calculus, often used to simplify complex integrals. It works by substituting a part of the integral with a new variable, making the expression easier to handle. In our problem, we used the substitution \( u = \sqrt{x} \), which means \( x = u^2 \) and \( dx = 2u \, du \). This substitution helps in reducing the original integral's complexity.

By converting \(\sqrt{x}\) into \(u\), the integral becomes more straightforward with polynomial expressions. Such a transition aids in handling different parts of the integral separately. Simpler integrals are usually more manageable and lead to easier solutions when conducting further calculus operations.

Partial Fraction Decomposition

Partial fraction decomposition is crucial when dealing with rational functions, especially during integration. Once a complicated rational function in the integrand is simplified by substitution, like the one we transformed into \( \frac{u - u^2}{1 + u} \), it needs further simplification.

The goal is to express these functions as a sum of simpler fractions. During decomposition, we separate the expanded form into simpler fractions. This involves writing the function as a sum, such as \( \int \frac{u}{1+u} \, du - \int u \, du \). These simpler fractions can each be integrated individually, often resulting in basic forms that are easily solvable.

Rational Functions

Rational functions are fractions where both the numerator and the denominator are polynomials. In calculus, integrating these functions can often seem daunting. However, methods like substitution and partial fraction decomposition simplify the process.

In our example, the integral originally involved a rational function \( \frac{1-\sqrt{x}}{1+\sqrt{x}} \). By using substitution, we transformed it into a function that could be handled through polynomial operations, \( \frac{u - u^2}{1 + u} \), thus enabling further simplification through partial fraction techniques. Understanding how to manipulate rational functions is crucial for solving more complex calculus problems.

Calculus Problem-Solving

Calculus often involves intricate problem-solving skills. Employing various techniques like substitution and partial fraction decomposition is part of a larger strategy to tackle complex tasks. These methods allow us to break down hard problems into simpler components.

Each step in solving a calculus problem serves as a building block toward the final solution. In our exercise, identifying the right substitution was the first critical step, followed by partial fraction decomposition to make the problem manageable. Problem-solving in calculus isn't just about finding the answer but understanding the process to derive simpler forms from complex integrals.