Proofs on integration of some standard functions part 5 - Integration - Diploma Maths - 2 | Summary and Q&A

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April 12, 2022
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Proofs on integration of some standard functions part 5 - Integration - Diploma Maths - 2

TL;DR

This video provides a step-by-step proof of the integral DX upon X square minus A square, which is equal to 1 upon 2A log X minus A upon X plus A plus C.

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Key Insights

  • ❎ The video provides a step-by-step proof of the integral DX upon X square minus A square.
  • ❎ The formula A square minus B square is used to solve the denominator.
  • 🥳 The integral is split into two parts and integrated separately.

Questions & Answers

Q: What is the formula used to solve the denominator in the proof?

The formula used is A square minus B square, which is equivalent to A minus B multiplied by A plus B.

Q: How is the integral separated into two parts in the proof?

The integral is split into 1 upon X minus A and 1 upon X plus A, which are then integrated separately.

Q: What is the derivative of X minus A in the proof?

The derivative of X minus A is 1.

Q: How is the final result obtained in the proof?

By integrating 1 upon X minus A and 1 upon X plus A separately and simplifying the expression, the final result of 1 upon 2A log of X minus A upon X plus A plus C is obtained.

Summary & Key Takeaways

  • The video presents a proof of the integral DX upon X square minus A square using similar steps from a previous video.

  • The denominator is solved using the formula A square minus B square, resulting in 1 upon 2A integral of 2A upon X minus A into X plus A DX.

  • The integral is then separated into two parts: 1 upon X minus A and 1 upon X plus A, which are then integrated to obtain the final result.

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