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# C3 integration :D watch

1. integrate:
x/(x-2)^2

answer: ln(x-1) - x/(x-1)

I got: x/x-1 - ln(x-1)

step by step method so I can see where I went wrong would be great and i'll give +ve rep
2. Integration by substitution. Know how to do that?

Let

AhH!. oops. that's my answer. latex confuses me
But I'm sure that's right, after looking voer it again
3. (Original post by maxfire)
Integration by substitution. Know how to do that?

Let

AhH!. oops. that's my answer. latex confuses me
But I'm sure that's right, after looking voer it again
That's definately not right...
I wouldn't do subsitiution as in the question it says use integration by parts
4. My method

u = x
du/dx = 1

dv/dx = 1/(x-2)^2 =(x-1)^-2
v=-1/x-1

answer = -x/(x-1) - integral of 1/x-2
=-x(x-1) + ln(x-1)

WOO i did it >.>
5. (Original post by SmileyGurl13)
My method

u = x
du/dx = 1

dv/dx = 1/(x-2)^2 =(x-1)^-2
v=-1/x-1

answer = -x/(x-1) - integral of 1/x-2
=-x(x-1) + ln(x-1)

WOO i did it >.>
Either I'm going mental, or there's something not quite right here.
6. Let and

I'm sure you can finish it from there.
Spoiler:
Show
So, the answer should be:

So, I disagree with the answer in the book or the answer given by the teacher.
7. out of interest, what are people typing up the solutions in? is it a maths program, or just word?
8. (Original post by edmhaslam)
out of interest, what are people typing up the solutions in? is it a maths program, or just word?
It's an inbuilt program in TSR called LateX. You can learn the basics here and then find some more codes here.
9. (Original post by GHOSH-5)
Let and

I'm sure you can finish it from there.
Ok, but how did change to ? Correct me if I'm wrong, but I'm quite sure that .
10. (Original post by meatball893)
Ok, but how did change to ? Correct me if I'm wrong, but I'm quite sure that .
Whoops, my eyes are going mad; I'll edit that post!
11. Three observations:

1. We've all (including me) forgotten our constants!
2. The book has a typo. If you replace 1s with 2s and imagine a constant it's all OK!
3. maxfire's solution was correct (though missing a constant). It is certainly not "definitely not right."

12. sweet
sorry not helping just wanted to try latex out :P
it's a bit of a chore though :/
13. If you put \displaystyle before \int it looks a lot nicer.

You get instead of .

And it's not so bad. You just have to get used to it.
14. (Original post by meatball893)
Either I'm going mental, or there's something not quite right here.
the dv/dx is gathered by the question as in it's the other part of it and then you have integrate it to work out v
15. (Original post by SmileyGurl13)
the dv/dx is gathered by the question as in it's the other part of it and then you have integrate it to work out v
I am familiar with integration by parts. I was merely questioning how 1/[(x-2)^2] became (x-1)^(-2).
16. (Original post by meatball893)
I am familiar with integration by parts. I was merely questioning how 1/[(x-2)^2] became (x-1)^(-2).
It's a mistake/typo in the answers you've been given.

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