Table of contents

0. Functions7h 52m
1. Limits and Continuity2h 2m
2. Intro to Derivatives1h 33m
3. Techniques of Differentiation3h 18m
4. Applications of Derivatives2h 38m
5. Graphical Applications of Derivatives6h 2m
6. Derivatives of Inverse, Exponential, & Logarithmic Functions2h 37m
7. Antiderivatives & Indefinite Integrals1h 26m
8. Definite Integrals4h 44m
9. Graphical Applications of Integrals2h 27m
- Area Between Curves
  1h 18m
- Introduction to Volume & Disk Method
  1h 8m
10. Physics Applications of Integrals 3h 16m
- Kinematics
  1h 13m
- Work
  2h 3m
11. Integrals of Inverse, Exponential, & Logarithmic Functions2h 34m
12. Techniques of Integration7h 39m
13. Intro to Differential Equations2h 55m
14. Sequences & Series5h 36m
15. Power Series2h 19m
- Introduction to Power Series
  1h 9m
- Taylor Series & Taylor Polynomials
  1h 10m
16. Parametric Equations & Polar Coordinates7h 58m

7. Antiderivatives & Indefinite Integrals

Indefinite Integrals

Calculus

7. Antiderivatives & Indefinite Integrals

Indefinite Integrals

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2:53 minutes

Problem 8.7.15

Textbook Question

7–40. Table look-up integrals Use a table of integrals to evaluate the following indefinite integrals. Some of the integrals require preliminary work, such as completing the square or changing variables, before they can be found in a table.
15. ∫ x / √(4x + 1) dx

Verified step by step guidance

Rewrite the integral in a form that is easier to match with a table of integrals. The given integral is ∫(x / √(4x + 1)) dx. Notice that the denominator involves a square root, so we will aim to simplify the expression.

Perform a substitution to simplify the square root. Let u = 4x + 1, which implies that du/dx = 4 or dx = du/4. Also, note that x = (u - 1)/4 from the substitution.

Substitute x and dx into the integral. The integral becomes ∫[((u - 1)/4) / √u] * (1/4) du. Simplify this expression to ∫[(u - 1) / (16√u)] du.

Split the integral into two simpler terms: ∫[(u / (16√u))] du - ∫[(1 / (16√u))] du. Simplify each term. The first term becomes (1/16)∫√u du, and the second term becomes (1/16)∫u^(-1/2) du.

Use the power rule for integration to evaluate each term. For the first term, ∫√u du = ∫u^(1/2) du = (2/3)u^(3/2). For the second term, ∫u^(-1/2) du = 2u^(1/2). Combine the results and substitute back u = 4x + 1 to express the solution 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

Indefinite integrals represent a family of functions whose derivative is the integrand. They are expressed without limits and include a constant of integration, typically denoted as 'C'. Understanding how to compute indefinite integrals is fundamental in calculus, as it allows for the determination of antiderivatives, which are essential in solving differential equations and analyzing functions.

Table of Integrals

A table of integrals is a reference tool that lists common integrals and their corresponding antiderivatives. It simplifies the process of finding integrals by providing ready-made solutions for frequently encountered functions. Familiarity with this table can significantly expedite the evaluation of integrals, especially when combined with techniques like substitution or completing the square.

Completing the Square

Completing the square is a method used to transform a quadratic expression into a perfect square trinomial. This technique is particularly useful in integration, as it can simplify the integrand, making it easier to apply integration techniques or look up in a table. For example, rewriting expressions in the form of (x + a)² helps in recognizing standard integral forms.

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