Integral of 1/(x^3+1) from 100 integrals | Summary and Q&A

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September 5, 2019
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blackpenredpen
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Integral of 1/(x^3+1) from 100 integrals

TL;DR

This video explains how to integrate the complex expression 1/(x^3+1) using partial fraction decomposition.

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Key Insights

  • ❓ Partial fraction decomposition is a useful technique for integrating rational functions with complex denominators.
  • 🧑‍🏭 Factoring the denominator is an important step before performing the partial fraction decomposition.
  • 😑 The constants in the decomposed fractions can be determined by equating coefficients in the original expression.

Transcript

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Questions & Answers

Q: What is the first step in integrating the expression 1/(x^3+1)?

The first step is to factor the denominator, which results in (x+1)(x^2-x+1).

Q: How is the partial fraction decomposition performed on the expression?

The partial fraction decomposition involves expressing the original expression as the sum of fractions with simpler denominators. In this case, it splits into A/(x+1) + (Bx + C)/(x^2-x+1).

Q: How are the constants A, B, and C determined in the partial fraction decomposition?

The constants A, B, and C are found by equating the coefficients of like powers of x in the original expression and the decomposed fractions. This can be done by using the cover-up method or setting up a system of equations.

Q: What is the final solution for the integral of 1/(x^3+1)?

The final solution involves integrating each term separately. The integral of 1/(x+1) can be found as the natural logarithm of the absolute value of (x+1). The integral of (Bx + C)/(x^2-x+1) can be found by completing the square and using inverse trigonometric functions.

Summary & Key Takeaways

  • The video discusses how to integrate the expression 1/(x^3+1) by first factoring the denominator into two factors: (x+1) and (x^2-x+1).

  • The partial fraction decomposition is then performed on the expression, resulting in the constants A, B, and C.

  • The integration is then performed on each separate term, leading to the final solution for the integral.

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