Thanks for the quick replies, I have 2 answers which I got not sure if there correct...

First Answer,

∫(y-1) dy = ∫x dx Mr F says: This is plainly wrong (at this level I do hope you understand why) and so all that follows from it will be wrong.

y^2 /2 - y = x^2 /2 + C1

Multiplied both sides by 2

y^2 - y + x^2 = C

= 12

Do I need to find 'C' in order to answer this question?

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My second answer is,

∫1/y-1 dy = ∫x dx Mr F says: Post #2 tells you that this is correct.

ln(y-1) = x^2 /2 + C1 Mr F says: Wrong. It is ln|y - 1| NOT ln(y - 1). There is a big difference that you should be aware of from basic calculus.

Mutliplied both sides by 2

ln(y-1) + x^2 = C Mr F says: If you multiply both sides by 2, and then subtract x^2, then surely you get 2 ln|y - 1| - x^2 = C. Although I don't know why you moved the x^2. It is better left as 2 ln|y - 1| = x^2 + C.

= 1.099 Mr F says: This has come out of nowhere and is meaningless without explanation.

Again do I need to find 'C' to answer this question or just leave it with at the integrated solution?

Thanks