Problem 23 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 15-34, use the method of substitution to find each of the following indefinite integrals. \(\int x\left(x^{2}+3\right)^{-12 / 7} d x\)

Step-by-Step Solution

Answer

The integral is \(-\frac{7}{10} (x^2 + 3)^{-5/7} + C\).

Step 1: Choose the Substitution

In the integration problem \(\int x(x^{2}+3)^{-12/7} \, dx\), notice the expression \(\left(x^2+3\right)\) inside the parentheses. We should choose this as our substitution variable:\(u = x^2 + 3\).

Step 2: Differentiate to Find \(du\)

Differentiate the substitution expression \(u = x^2 + 3\) with respect to \(x\).\[\frac{du}{dx} = 2x \]Then solve for \(dx\):\[du = 2x \, dx \Rightarrow dx = \frac{du}{2x}\]

Step 3: Substitute in the Integral

Substitute \(u\) and \(dx\) into the original integral.\[\int x(x^2+3)^{-12/7} \, dx = \int x(u)^{-12/7} \cdot \frac{du}{2x}\]

Step 4: Simplify the Integral

Cancel the \(x\) terms in the integral.\[\int x(u)^{-12/7} \cdot \frac{du}{2x} = \int \frac{1}{2}u^{-12/7} \, du\]This simplifies to:\[\frac{1}{2}\int u^{-12/7} \, du\]

Step 5: Perform the Integration

Integrate \(\frac{1}{2} \int u^{-12/7} \, du\) using the power rule for integration.\[\frac{1}{2} \left( \frac{u^{-12/7+1}}{-12/7+1} \right) + C\]Simplify the exponent \(-12/7 + 1 = -5/7\):\[\frac{1}{2} \left( \frac{u^{-5/7}}{-5/7} \right) + C\]Simplify further:\[\frac{1}{2} \times \left( -\frac{7}{5} u^{-5/7} \right) + C\]This gives:\[-\frac{7}{10} u^{-5/7} + C\]

Step 6: Substitute Back for \(x\)

Substitute \(u = x^2 + 3\) back into the expression.\[- \frac{7}{10} (x^2 + 3)^{-5/7} + C\]Now our solution is in terms of \(x\).

Key Concepts

Method of Substitution

The Method of Substitution is an essential technique in calculus for evaluating integrals. It simplifies complex integrals by introducing a new variable. This method is particularly useful when dealing with functions inside a composite form, like nested functions.Here's how it works in a nutshell:

• First, identify a part of the integrand that can be rewritten as a single variable (often something inside parentheses).
• Set this part equal to a new variable, say \( u \).
• Differentiate to find \( du \), which represents the change in \( u \).
• Replace both the chosen part of the integrand and \( dx \) with \( u \) and \( du \) in the integral.
• This substitution often turns the integral into a much simpler form, making it easier to evaluate.

This method allows us to integrate functions that appear complicated at first glance. It changes the perspective by transforming the variable of integration temporarily, simplifying the integral.

Power Rule for Integration

The Power Rule for Integration is one of the foundational tools used in calculus. It helps simplify the process of finding the integral of a function with a variable raised to a power. The Power Rule states:\[ \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \] This is applicable as long as \( n ≠ -1 \). The key steps involve:

• Adding one to the exponent \( n \) of the variable \( x \).
• Dividing the new term by the new exponent \( n+1 \).
• Don't forget to add \( C \), the constant of integration, since we are finding an indefinite integral.

In our exercise, once we have simplified the integral using substitution, we apply the Power Rule to find the integral of \( u^{-12/7} \). The simplicity of the Power Rule is why it is a favored method in calculus, helping to quickly solve problems by reducing the complexity of differentiation inverses.

Integration by Substitution

Integration by Substitution is a method deeply linked to the Method of Substitution but focuses on transforming the integrals to make them easier to work with. By substituting parts of the integral with simpler forms, it aligns the integral in a way that often takes advantage of basic integration rules.Here are the core steps in the integration by substitution process:

• Choose a substitution \( u = g(x) \), where \( g(x) \) is part of the integrand that complicates integration.
• Calculate \( du \), which involves differentiating \( g(x) \) with respect to \( x \).
• Replace all instances of \( x \) in the integral with \( u \) and \( dx \) with \( \frac{du}{g'(x)} \). This transforms the integral to a simpler form.
• Find the integral of the transformed expression in terms of \( u \).
• Finally, substitute back the original expression for \( u \) to express the integral in terms of \( x \).

This method, employed in the provided solution, is especially useful for integrands with composite functions, as it translates the problem into something doable using more straightforward techniques.

Calculus Techniques

In calculus, there are several techniques for solving integrals. These techniques vary in complexity and applicability, but they all aim to simplify integration, turning a complex problem into a more manageable one.Some widely used techniques include:

• Substitution: As explained, it simplifies through variable replacement.
• Integration by Parts: Useful when the product of functions is involved, using the identity \( \int u \, dv = uv - \int v \, du \).
• Partial Fraction Decomposition: Breaks down more complex rational expressions into simpler fractions.
• Trigonometric Substitution: Applies when integrals involve square roots of expressions like \( a^2 - x^2 \).

Choosing the right calculus technique depends on the nature of the integrand. Recognizing patterns and structures within an integral can guide you to the most appropriate method. The goal is always to transform challenging functions into forms where basic integration rules can be applied efficiently. The choice of technique can dramatically affect the ease with which the integral is solved, and practiced understanding of these methods is key to mastering the art of integration.