Question

Use the given substitution to convert the integral to an integral of a rational function, then evaluate. \(\int \frac{1}{\sqrt{x}+\sqrt[3]{x}} d x ; x=u^{6}\)

Short answer

\(2 \sqrt{x} - 3 \sqrt[3]{x} + 6 \sqrt[6]{x} - 6 \ln|\sqrt[6]{x} + 1| + C\)

Step-by-step solution

Apply the Substitution

Given the substitution \(x = u^6\), rewrite the original integral in terms of \(u\). First, calculate the derivatives: \(dx = 6u^5 du\). Substitute \(x = u^6\), \(\sqrt{x} = u^3\), and \(\sqrt[3]{x} = u^2\) into the integral.

Simplify the Integral

Replace \(\sqrt{x}\) and \(\sqrt[3]{x}\) in the integral: \[\int \frac{1}{\sqrt{x} + \sqrt[3]{x}} \, dx = \int \frac{1}{u^3 + u^2} \, 6u^5 du\] Now simplify: \[\int \frac{6u^5}{u^3 + u^2} \, du\]

Simplify the Rational Function

Rewrite the expression by factoring out the common terms: \[6 \int \frac{u^5}{u^2(u + 1)} \, du = 6 \int \frac{u^3}{u + 1} \, du\]

Perform Long Division

Divide \(u^3\) by \(u+1\) using polynomial long division to simplify the integrand: \[u^3 \div (u+1) = u^2 - u + 1 - \frac{1}{u+1}\]

Integrate Each Term

Now separately integrate each term from the division: \[6 \int (u^2 - u + 1 - \frac{1}{u+1}) \, du = 6 \left( \frac{u^3}{3} - \frac{u^2}{2} + u - \ln|u+1| \right) + C\]

Back-substitute the Original Variable

Reverse the substitution \(u = x^{1/6}\) to express the integral in terms of \(x\): \[6 \left( \frac{(x^{1/6})^3}{3} - \frac{(x^{1/6})^2}{2} + x^{1/6} - \ln|x^{1/6}+1| \right) + C\]Simplify: \[2 x^{1/2} - 3 x^{1/3} + 6 x^{1/6} - 6 \ln|x^{1/6} + 1| + C\]

Write the Final Answer

Therefore, the integral evaluates to:\[2 \sqrt{x} - 3 \sqrt[3]{x} + 6 \sqrt[6]{x} - 6 \ln|\sqrt[6]{x} + 1| + C\]

Key concepts

Substitution Method

The substitution method is a common technique used in calculus to simplify the integration process. It involves replacing one variable with another to transform an integral into a simpler form. In our exercise, we use the substitution \( x = u^6 \) to reframe the original integral. This substitution turns the complex powers of \( x \) into manageable powers of \( u \).
This method requires computing the derivative of the substituted variable. Here, we differentiate \( x = u^6 \) to get \( dx = 6u^5 \, du \). Once the substitution is applied, you replace all \( x \)-related expressions in the integral with their \( u \) equivalents, simplifying the integration process.

• The main goal is to convert the complex expression into one involving basic polynomials or rational functions.
• After integration, you must replace \( u \) with its expression in terms of \( x \) to return to the variable of the original problem.

This method is highly effective for integrals that resemble compositions of functions or have nested radicals or powers.

Rational Function Integration

Integrating rational functions is a crucial part of calculus, involving the integration of fractions where both the numerator and the denominator are polynomials. In our example, after substitution, we end up with the integral \( \int \frac{6u^5}{u^3 + u^2} \, du \). This is a form of a rational function because it's a polynomial divided by another polynomial.
Simplification precedes integration, often involving factoring and polynomial long division to break down the expression. Factoring helps to identify potential simplifications such as canceling common terms in the numerator and the denominator.

• Understanding and manipulating polynomial structures are essential for successful integration.
• Rational function integration often precedes applying specialized techniques like partial fraction decomposition unless a simpler factorization is easily accomplished.

Rational function integration is especially valuable in various fields, such as physics and engineering, where systems are often modeled by polynomial equations.

Polynomial Long Division

Polynomial long division is similar to numerical long division but involves dividing polynomials instead of numbers. It is a technique used to simplify a complex rational function into a more integrable form. In the example, we perform long division on \( u^3 \) by \( u+1 \) to simplify the integrand.
Through this division, we express \( u^3 \) as \( u^2 - u + 1 - \frac{1}{u+1} \), transforming the integral into a form that is easier to work with. Each term from the division can be independently integrated, making the process straightforward.

• The quotient from polynomial division often reveals terms that integrate directly, such as simple powers of \( u \).
• Any remainder is managed with further integration methods, like substitution or partial fraction decomposition.

Employing polynomial long division strips down complicated expressions into manageable pieces, paving the way for effective calculus integration.