Question

Use substitution to convert the integrals to integrals of rational functions. Then use partial fractions to evaluate the integrals. \(\int \frac{d x}{\sqrt{x}+\sqrt[4]{x}}\)

Short answer

The evaluated integral is \( 4 \ln \left( \frac{\sqrt[4]{x}}{\sqrt[4]{x} + 1} \right) + C \).

Step-by-step solution

Find a Suitable Substitution

To simplify the integral, use the substitution \( u = \sqrt[4]{x} \). This implies \( x = u^4 \). Differentiate to find \( dx \), leading to \( dx = 4u^3 \, du \). Substitute these values into the integral.

Substitute and Simplify the Integral

Replace \( \sqrt{x} \) with \( u^2 \) and \( \sqrt[4]{x} \) with \( u \) in the integrand. The integral becomes \( \int \frac{4u^3 \, du}{u^2 + u} \). Simplify this to \( 4 \int \frac{u^2}{u(u+1)} \, du \).

Simplify Further and Use Partial Fractions

Simplify the integrand to \( 4 \int \frac{u^2}{u(u+1)} \, du = 4 \int \frac{1}{u} \, du - 4 \int \frac{1}{u+1} \, du \). Use partial fractions to separate the terms.

Integrate Using Basic Integration Rules

Integrate each term separately: \( 4 \int \frac{1}{u} \, du \) becomes \( 4 \ln|u| \) and \( -4 \int \frac{1}{u+1} \, du \) becomes \(-4 \ln|u+1| \). Combine these results to form \( 4 \ln|u| - 4 \ln|u+1| + C \).

Back-Substitute and Simplify Result

Substitute back \( u = \sqrt[4]{x} \), giving \( 4 \ln|\sqrt[4]{x}| - 4 \ln|\sqrt[4]{x} + 1| + C \). This simplifies to \( 4 (\ln|\sqrt[4]{x}| - \ln|\sqrt[4]{x} + 1|) + C \). So the final answer is \( 4 \ln \left( \frac{\sqrt[4]{x}}{\sqrt[4]{x} + 1} \right) + C \).

Key concepts

Substitution Method

The substitution method is essential in calculus integration, especially when dealing with complex integrals. It involves changing the variable of integration to simplify the integral, making it easier to solve. In our exercise, we used the substitution \( u = \sqrt[4]{x} \). This transformation is strategic because it turns complicated expressions into simpler forms. By setting \( u = \sqrt[4]{x} \), we express \( x \) as \( u^4 \) and differentiate to find \( dx = 4u^3 \, du \). This transformation allows us to substitute \( \sqrt{x} \) with \( u^2 \) and the \( \sqrt[4]{x} \) with \( u \). As a result, the original integral \( \int \frac{dx}{\sqrt{x}+\sqrt[4]{x}} \) is transformed into a more manageable form: \( \int \frac{4u^3 \, du}{u^2 + u} \). Using substitution is a powerful tool in integration as it can simplify expressions significantly, allowing for easier computation and analysis.

Partial Fraction Decomposition

Partial fraction decomposition is a method used to break down rational expressions into simpler fractions. This technique is particularly useful when addressing integrals that result from the substitution method. Once we substituted and simplified the expression in our exercise, it became \( 4 \int \frac{u^2}{u(u+1)} \, du \). To integrate this form, we decompose it using partial fractions. The goal is to express the integrand as \( \frac{1}{u} - \frac{1}{u+1} \). This decomposition makes integration straightforward, allowing us to integrate each term separately using basic integration rules. The computation simplifies the integral into two parts, \( 4 \int \frac{1}{u} \, du - 4 \int \frac{1}{u+1} \, du \), which are easy to integrate. Understanding partial fraction decomposition is crucial, as it breaks complex rational expressions into simpler parts that are easier to integrate.

Rational Functions

Rational functions are quotients of polynomials and appear frequently in calculus problems, particularly in integration. Recognizing properties of rational functions facilitates methods like substitution and partial fraction decomposition. In our exercise, the rational function \( \frac{4u^3}{u^2 + u} \) was simplified down to a format amenable for integration through substitution and partial fraction decomposition. This function type has specific patterns and properties that can be exploited to simplify integration. Dealing with rational functions often leads to decompositions or transformations into expressions like \( \frac{1}{u} \) and \( \frac{1}{u+1} \), whose antiderivatives are commonly known. Understanding rational functions not only aids in integration but also enhances problem-solving capabilities in calculus.Familiarity with rational functions enhances analysis and problem-solving skills, making it easier to apply techniques like substitution and partial fraction decomposition effectively.