Integration of Linear and Quadratic Expression problem No. 7 - Integration - Diploma Maths - 2 | Summary and Q&A

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April 12, 2022
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Integration of Linear and Quadratic Expression problem No. 7 - Integration - Diploma Maths - 2

TL;DR

Solving a problem involving integration of a linear expression divided by a quadratic expression.

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Key Insights

  • πŸ˜‘ The problem involves integrating a linear expression divided by a quadratic expression.
  • ☺️ The values of a and B can be determined by comparing the coefficients of X in the given expression.
  • πŸ”™ After determining a and B, the integral can be simplified and solved using standard integration formulas.

Transcript

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Questions & Answers

Q: What is the problem in this video about?

The problem is about integrating the expression X/(X^2 - 6X + 13).

Q: How can the value of a and B be determined in the problem?

By substituting X as a into dX/(dX^2 - 6X + 13) + B, the values of a and B can be determined by comparing the coefficients of X.

Q: How is the integral simplified further after determining the values of a and B?

After determining the values of a and B, the integral can be written as separate terms and simplified using standard integration formulas.

Q: What is the final solution to the problem?

The final solution is AI = 1/2 * log(X^2 - 6X + 13) + 3/2 * tan inverse (X - 3/2) + C.

Summary & Key Takeaways

  • The video tutorial explains how to solve a problem that involves integrating an expression of the form X/(X^2 - 6X + 13).

  • The linear expression in this problem is X and the quadratic expression is X^2 - 6X + 13.

  • By substituting X as a into dX/(dX^2 - 6X + 13) + B, the values of a and B can be determined.

  • Once the values of a and B are known, the integral can be simplified further and solved.

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