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# integrating a curve watch

1. I've been asked to integrate/"find general solution" of
(x(y^(2) + 1)) *((x-1)y)^(-1)

I'm struggling a lot with these last few chapters... Chapter 7 Differential Equations and 8 Curves defined Implicitly. Many of my class mates are stuggling too but that's little consolation :/ I don't really know how to start... was barely able to do Chapter 7 last week.
2. (Original post by one-final-hit)
I've been asked to integrate/"find general solution" of
(x(y^(2) + 1)) *((x-1)y)^(-1)

I'm struggling a lot with these last few chapters... Chapter 7 Differential Equations and 8 Curves defined Implicitly. Many of my class mates are stuggling too but that's little consolation :/ I don't really know how to start... was barely able to do Chapter 7 last week.
Is that

?

If so, have you tried to separate the variables?

4. Yeh That was helpful; thank you to both of you.
I've come to this now:
1/2 * Ln(y^(2)+1) = x - 1 - ln(x-1) + C

Not sure whether the right hand side is correct, and now what to do :/

If the right hand side is correct, I've taken it down to:
ln(y^(2)+1) = 2(x-1)-ln((x-1)^(2)) +c
5. (Original post by one-final-hit)
Yeh That was helpful; thank you to both of you.
I've come to this now:
1/2 * Ln(y^(2)+1) = x - 1 - ln(x-1) + C

Not sure whether the right hand side is correct, and now what to do :/

If the right hand side is correct, I've taken it down to:
ln(y^(2)+1) = 2(x-1)-ln((x-1)^(2)) +c
Use integration by parts on the right hand side, and you should get (x-1)(ln(x-1)) instead

The solution should be ln(y^(2)+1) = 2(x-1)(ln(x-1)) + C
6. (Original post by jameswhughes)
Use integration by parts on the right hand side, and you should get (x-1)(ln(x-1)) instead

The solution should be ln(y^(2)+1) = 2(x-1)(ln(x-1)) + C
Can you explain how you used integration by parts because IMO that's the slower route I chose not to take simply because there was too much maths involved to justify it... im stuck either differentiating (x-1)^1 or integrating Ln(x-1).

I figured out my mistake with u substitution in that I put x = u-1 instead of u+1, but my answer is still different to yours..
= x-1+ln(x-1) + c

Close to giving up seeing as I'm the only person in my class to have got this far...

[edit: usubs]

S x(x-1)^(-1) dx Where x-1=u, x=u+1, dx=1*du S= integral sign
S ((u+1)/u) * 1 du
S (u/u)+(1/u) du
S 1+(1/u) du
= [u +lnu +c]
=x-1+ln(x-1) +c
7. (Original post by one-final-hit)
Can you explain how you used integration by parts because IMO that's the slower route I chose not to take simply because there was too much maths involved to justify it... im stuck either differentiating (x-1)^1 or integrating Ln(x-1).

I figured out my mistake with u substitution in that I put x = u-1 instead of u+1, but my answer is still different to yours..
= x-1+ln(x-1) + c

Close to giving up seeing as I'm the only person in my class to have got this far...
8. Ah ok, no problems
mmm looking at the answer in the back of the book to try cook up a methd:
y=+-sqrt((k(x-1)^(2) * e^(2x)) - 1)

Can anyone tell me where that e^(2x) is coming from?
[It's appeared once before in Chapter 7 Misc 7 Q8, and I didn't know what to do; looked in the book and it wasnt in that chapter so left it empty; teacher afaik, didn't explain it]
9. (Original post by one-final-hit)
Ah ok, no problems
mmm looking at the answer in the back of the book to try cook up a methd:
y=+-sqrt((k(x-1)^(2) * e^(2x)) - 1)

Can anyone tell me where that e^(2x) is coming from?
[It's appeared once before in Chapter 7 Misc 7 Q8, and I didn't know what to do; looked in the book and it wasnt in that chapter so left it empty; teacher afaik, didn't explain it]
I think they have just multiplied by e, as [k1]e^(2x-1)=[k2]e^2x if the k2 is just another constant.

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