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Ch. 4 - Applications of Derivatives
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 4 - Applications of DerivativesProblem 4.PE.81
Chapter 4, Problem 4.PE.81

Finding Indefinite Integrals
Find the indefinite integrals (most general antiderivatives) in Exercises 73–88. You may need to try a solution and then adjust your guess. Check your answers by differentiation.

∫ 𝓍³ (1 + 𝓍⁴ )⁻¹/⁴ d𝓍

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1
Identify the integral to solve: \(\int x^{3} (1 + x^{4})^{-\frac{1}{4}} \, dx\).
Look for a substitution that simplifies the integral. Notice that the expression inside the parentheses is \(1 + x^{4}\), and its derivative involves \(x^{3}\). This suggests using the substitution \(u = 1 + x^{4}\).
Compute the differential \(du\) by differentiating \(u\) with respect to \(x\): \(du = 4x^{3} \, dx\). From this, solve for \(x^{3} \, dx\) to express it in terms of \(du\): \(x^{3} \, dx = \frac{du}{4}\).
Rewrite the integral in terms of \(u\) using the substitution: replace \(x^{3} \, dx\) with \(\frac{du}{4}\) and \((1 + x^{4})^{-\frac{1}{4}}\) with \(u^{-\frac{1}{4}}\). The integral becomes \(\int u^{-\frac{1}{4}} \cdot \frac{du}{4}\).
Integrate the new integral with respect to \(u\): \(\frac{1}{4} \int u^{-\frac{1}{4}} \, du\). Use the power rule for integration, which states \(\int u^{n} \, du = \frac{u^{n+1}}{n+1} + C\) for \(n \neq -1\). After integrating, substitute back \(u = 1 + x^{4}\) to express the answer in terms of \(x\).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Indefinite Integrals and Antiderivatives

An indefinite integral represents the most general antiderivative of a function, including an arbitrary constant of integration. It reverses differentiation, finding a function whose derivative matches the integrand. Understanding this concept is essential for solving integrals and verifying solutions by differentiation.
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Substitution Method

The substitution method simplifies integration by changing variables to transform a complex integral into a more manageable form. Typically, a part of the integrand is set as a new variable, allowing the integral to be rewritten in terms of this variable. This technique is especially useful when the integrand contains composite functions.
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Power Rule for Integration

The power rule states that the integral of x^n (for n β‰  -1) is (x^(n+1))/(n+1) plus a constant. This rule is fundamental for integrating polynomial expressions and powers of variables. Recognizing when to apply this rule after substitution helps in efficiently solving integrals involving powers.
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Related Practice
Textbook Question

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Solve the initial value problems in Exercises 89–92.


dy/dx = (𝓍 + 1/𝓍)Β² , y(1)= 1

Textbook Question

Initial Value Problems

Solve the initial value problems in Exercises 89–92.

d^3 r/dt^3 = - cos t; r''(0) = r'(0) = 0 , r(0) = -1

Textbook Question

Finding Indefinite Integrals

Find the indefinite integrals (most general antiderivatives) in Exercises 73–88. You may need to try a solution and then adjust your guess. Check your answers by differentiation.

∫ sec² s/10 ds

Textbook Question

Finding Indefinite Integrals

Find the indefinite integrals (most general antiderivatives) in Exercises 73–88. You may need to try a solution and then adjust your guess. Check your answers by differentiation.

∫ cosΒ³ 𝓍/2 d𝓍

Textbook Question

Finding Indefinite Integrals

Find the indefinite integrals (most general antiderivatives) in Exercises 73–88. You may need to try a solution and then adjust your guess. Check your answers by differentiation.

__

∫ ( 3√ t + 4/t² ) dt

Textbook Question

Finding Indefinite Integrals

Find the indefinite integrals (most general antiderivatives) in Exercises 73–88. You may need to try a solution and then adjust your guess. Check your answers by differentiation.

∫ sec θ/3 tan θ/3 dθ