Question

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

Short answer

\( \frac{3}{2} \ln \left| \frac{\sqrt[3]{t}-1}{\sqrt[3]{t}+1} \right| + C. \)

Step-by-step solution

Apply the Substitution

We are given the substitution \( t = x^3 \). Thus, we need to express \( dt \) in terms of \( dx \). Differentiating \( t = x^3 \) with respect to \( x \) gives \( dt = 3x^2 dx \). This substitution will help rewrite the integral in terms of \( x \).

Rewrite the Integral in Terms of x

Using the substitution \( t = x^3 \) and \( dt = 3x^2 dx \), substitute these into the given integral:\[ \int \frac{1}{t- oot{3}{t}} \, dt \rightarrow \int \frac{1}{x^3-x} \, (3x^2 \, dx). \] Thus, the integral becomes:\[ 3 \int \frac{x^2}{x^3-x} \, dx. \]

Simplify the Integrand

Factor the denominator:\[ x^3-x = x(x^2-1) = x(x-1)(x+1). \] The integral becomes:\[ 3 \int \frac{x^2}{x(x-1)(x+1)} \, dx = 3 \int \frac{x}{(x-1)(x+1)} \, dx. \]

Perform Partial Fraction Decomposition

Write \( \frac{x}{(x-1)(x+1)} \) as the sum of partial fractions:\[ \frac{x}{(x-1)(x+1)} = \frac{A}{x-1} + \frac{B}{x+1}. \]Multiply through by the common denominator to solve for \( A \) and \( B \):\[ x = A(x+1) + B(x-1). \]

Solve for the Constants A and B

Equating coefficients from both sides:1. At \( x = 1 \), \( 1 = A(1+1) + B(0) \Rightarrow A = \frac{1}{2} \).2. At \( x = -1 \), \( -1 = 0 + B(-1-1) \Rightarrow B = -\frac{1}{2} \).Thus, \( \frac{x}{(x-1)(x+1)} = \frac{1/2}{x-1} - \frac{1/2}{x+1} \).

Integrate Each Fraction

Integrate the expression:\[ 3 \int \left( \frac{1/2}{x-1} - \frac{1/2}{x+1} \right) \, dx = \frac{3}{2} \int \frac{1}{x-1} \, dx - \frac{3}{2} \int \frac{1}{x+1} \, dx. \]This results in:\[ \frac{3}{2} \ln |x-1| - \frac{3}{2} \ln |x+1| + C. \]

Simplify the Logarithmic Expression

Combine the logarithms:\[ \frac{3}{2} \left( \ln |x-1| - \ln |x+1| \right) = \frac{3}{2} \ln \left| \frac{x-1}{x+1} \right| + C. \]

Substitute Back to t

Recall \( x = \sqrt[3]{t} \). Substitute back:\[ \frac{3}{2} \ln \left| \frac{\sqrt[3]{t}-1}{\sqrt[3]{t}+1} \right| + C. \]

Key concepts

Substitution Method

The substitution method in integral calculus is a powerful technique. It is used to simplify complex integrals by transforming the variable of integration. In general, you might find dealing with a direct integral cumbersome, and substituting offers a way to make it more manageable.
For example, consider substituting a variable like \( t \rightarrow x^3 \). This transformation allows you to change the measure of integration from \( dt \) to \( dx \). You achieve this by differentiating your substitution equation, giving you \( dt = 3x^2 \, dx \). This substitution simplifies the original integral into one with a more familiar and approachable form, often turning an irrational function into a rational one.
By rewriting the integral in terms of \( x \), subsequent manipulations of the integral become feasible and often significantly easier. Substitution is akin to changing perspectives, providing a fresh view on the integral that sometimes reveals clearer solutions.

Partial Fraction Decomposition

Partial fraction decomposition is a technique used in integral calculus to break down complex rational expressions. This method expresses a complicated fraction as a sum of simpler fractions, making integration straightforward.
Given a fraction like \( \frac{x}{(x-1)(x+1)} \), the idea is to represent it as \( \frac{A}{x-1} + \frac{B}{x+1} \). The unknowns \( A \) and \( B \) are constants we solve for, making the expression easier to handle. This requires setting up an equation that equates the original rational expression with the decomposed form multiplied back out.
Solving for these constants usually involves plugging in strategic values for the variable to find \( A \) and \( B \). Once decomposed, each fraction can be integrated separately, turning an intimidating calculus task into a simpler one. This method is powerful for handling integrals involving polynomial divisions.

• Essential for simplifying complex expressions.
• Useful in changing denominators into easier formats.
• Transforms polynomial fractions into linear integrations.

Logarithmic Functions

Logarithmic functions are deeply connected with natural patterns and growths, making them essential in calculus. When integrating expressions, especially those acquired through partial fractions, logarithmic functions often arise.
For instance, integrating forms like \( \frac{1}{x-a} \) results in \( \ln |x-a| \). Combining logarithmic expressions through subtraction, such as \( \ln |x-1| - \ln |x+1| \), can be condensed using logarithmic identities. Specifically, this can be converted into a single expression: \( \ln \left| \frac{x-1}{x+1} \right| \).
Understanding how to manipulate these types of expressions is crucial for expressing the final form of your solution cleanly. Here, logarithmic identities play a vital role, simplifying complex expressions into neat, concise forms.

Rational Functions

Rational functions are quotients of polynomials. These functions are prevalent in calculus because they represent many real-world phenomena. Integrating rational functions is a common task, where techniques like substitution and partial fraction decomposition shine.
Rational functions like \( \frac{x}{(x^3-x)} \) may initially seem formidable due to the polynomial terms in their numerators and denominators. Factoring the denominator into \( x(x^2-1) \) or similar components simplifies the problem. The simplification makes it easier to perform algebraic manipulations which are necessary for the decomposition.
While working through rational functions, it is important to consider if the expression can be simplified further or if other calculus techniques can be applied. Overall, understanding the behavior of these functions and their properties allows you to tackle their integrals methodically and efficiently.

• Includes polynomial expressions in both numerator and denominator.
• Factoring helps in reducing complexity.
• Essential for modeling and analysis in various fields.