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.

Transcript

click the bell icon to get latest videos from equator hello friends in this video we are going to see problem number 7 which is based on integral linear expression upon quadratic expression let us start integral X upon X square minus 6x plus 13 DX here you can see the linear expression is X and a quadratic expression is X square minus 6x plus 13 th... Read More

Key Insights

• 😑 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.
• 🔙 After determining a and B, the integral can be simplified and solved using standard integration formulas.

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.