Indefinite Integrals

[cryo]

Question:



Their solution:






My solution:


Apologies for the messiness, and dookie quality.

But I'd be thankful if someone could help me understand why I got different values for p and q through my method of solution.
Is it because my simplification was flawed?

 

[Deleted member 4993]

Question:
View attachment 21921


Their solution:

View attachment 21922




My solution:
View attachment 21923

Apologies for the messiness, and dookie quality.

But I'd be thankful if someone could help me understand why I got different values for p and q through my method of solution.
Is it because my simplification was flawed?

Look at the first line of response
You write

\(\displaystyle \frac{P}{2x^2} \ \to \ P \left(2 \right) \ x^{-2}\) ...... small but fatal mistake.

It should be:

\(\displaystyle \frac{P}{2x^2} \ \to \ P \left( \frac{1}{2}\right) x^{-2} \ \to \ P \left( 2^{-1}\right) x^{-2} \)

There could be more problems but I did not check for those if present.

 

[cryo]

I thank you, Subhotosh!