Proofs on integration of some standard functions part 5  Integration  Diploma Maths  2  Summary and Q&A
TL;DR
This video provides a stepbystep 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.
Key Insights
 ❎ The video provides a stepbystep 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.