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

Antiderivatives

Calculus

7. Antiderivatives & Indefinite Integrals

Antiderivatives

Previous problemNext problem

2:47 minutes

Problem 4.9.21

Textbook Question

Finding antiderivatives. Find all the antiderivatives of the following functions. Check your work by taking derivatives.

g(s) = 1 / (s² + 1)

Verified step by step guidance

Step 1: Recognize that the function g(s) = 1 / (s² + 1) resembles the derivative of the arctangent function. Recall the formula for the derivative of arctan(x): d/dx [arctan(x)] = 1 / (x² + 1).

Step 2: Use the antiderivative rule to conclude that the integral of g(s) = 1 / (s² + 1) is arctan(s) + C, where C is the constant of integration.

Step 3: Write the general form of the antiderivative: G(s) = arctan(s) + C.

Step 4: To verify your work, take the derivative of G(s). Using the derivative rule for arctan(s), d/ds [arctan(s)] = 1 / (s² + 1), and the derivative of a constant is 0.

Step 5: Confirm that the derivative of G(s) matches the original function g(s). Since d/ds [arctan(s) + C] = 1 / (s² + 1), the solution is correct.

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Key Concepts

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

Antiderivatives

An antiderivative of a function is another function whose derivative is the original function. In calculus, finding antiderivatives is essential for solving problems related to integration. The process often involves recognizing patterns or applying integration techniques, such as substitution or integration by parts.

Integration Techniques

Integration techniques are methods used to find antiderivatives of functions that may not be straightforward. Common techniques include substitution, integration by parts, and partial fraction decomposition. Understanding these methods is crucial for effectively solving integrals, especially for more complex functions.

Verification by Differentiation

Verification by differentiation involves taking the derivative of the antiderivative found to ensure it matches the original function. This step is important for confirming the correctness of the antiderivative. It reinforces the fundamental theorem of calculus, which connects differentiation and integration.