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# Help with C3 Integration question watch

1. Integrate x/x-2 in the range from 3 to 4.
So far I've only been taught how to integrate via substation so I thought u = x-2 and tried doing the usual but it doesn't seem to work out. Is there an alternative method?
2. (Original post by MrToodles4)
Integrate x/x-2 in the range from 3 to 4.
So far I've only been taught how to integrate via substation so I thought u = x-2 and tried doing the usual but it doesn't seem to work out. Is there an alternative method?
The substitution should work - post your working, and someone can check it out.

Alternatively, you can just perform the polynomial/synthetic division, if you've covered it (don't know what module that is), and then integrate.
3. (Original post by MrToodles4)
Integrate x/x-2 in the range from 3 to 4.
So far I've only been taught how to integrate via substation so I thought u = x-2 and tried doing the usual but it doesn't seem to work out. Is there an alternative method?
You could also write the fraction x/(x-2) as 1 + 2/(x-2). If you subtract 2 and then add 2 to the numerator it might be easier to see why.
4. (Original post by akaash13)
You could also write the fraction x/(x-2) as 1 + 2/(x-2). If you subtract 2 and then add 2 to the numerator it might be easier to see why.
How did you get this?
5. If you post your working out or what you have been doing, we can see where the confusion is and can then help you.
6. (Original post by MrToodles4)
How did you get this?
x/{x - 2} is the same as {x - 2 + 2}/{x - 2}
7. (Original post by MrToodles4)
How did you get this?
$\int \frac{x-2+2}{x-2} = \int \frac{(x-2)+2}{x-2} = \int \frac{(x-2)}{x-2} + \frac{2}{x-2} = \int 1 + \frac{2}{x-2}$

Left out the dx because there was not enough space. :/
The u-substitution should also work. Just remember to change the limits and you should get (u+2)/u = 1 + 2/u.
8. (Original post by ghostwalker)
The substitution should work - post your working, and someone can check it out.

Alternatively, you can just perform the polynomial/synthetic division, if you've covered it (don't know what module that is), and then integrate.
So I wrote out the equation as x(x-2)^-1. And let u=x-2. Therefore du/dx = 1. dx/du = 1.

so x*u^-1. to get x in terms of u i did u+2=x
So (u+2)*u^-1. I multiplied out the bracket to get 1+2u^-1.
If this is integrated this is the same as u+2lnu. And then I put in the new ranges from intergral u=1 and u= 2. (originally x=3 and x=4) so 2+2ln2 - 1+2ln1 however this does not give me the correct answer of 1+2ln2. Does the 1 not need to be integrated into u? is that the mistake Im making here?
9. (Original post by MrToodles4)
So I wrote out the equation as x(x-2)^-1. And let u=x-2. Therefore du/dx = 1. dx/du = 1.

so x*u^-1. to get x in terms of u i did u+2=x
So (u+2)*u^-1. I multiplied out the bracket to get 1+2u^-1.
If this is integrated this is the same as u+2lnu. And then I put in the new ranges from intergral u=1 and u= 2. (originally x=3 and x=4) so 2+2ln2 - 1+2ln1 however this does not give me the correct answer of 1+2ln2. Does the 1 not need to be integrated into u? is that the mistake Im making here?
2+2ln2 - 1+2ln1

1+2ln2

These are the same thing since ln(1) = 0.

For future questions please post all your working at the start so that we know where you're stuck.
10. Note: (a+b) - (c+d) = a + b - c - d, (note the bold emphasis).
11. (Original post by MrToodles4)
So I wrote out the equation as x(x-2)^-1. And let u=x-2. Therefore du/dx = 1. dx/du = 1.

so x*u^-1. to get x in terms of u i did u+2=x
So (u+2)*u^-1. I multiplied out the bracket to get 1+2u^-1.
If this is integrated this is the same as u+2lnu. And then I put in the new ranges from intergral u=1 and u= 2. (originally x=3 and x=4) so 2+2ln2 - 1+2ln1 however this does not give me the correct answer of 1+2ln2. Does the 1 not need to be integrated into u? is that the mistake Im making here?
let u = x-2 , x = u +2
du/dx = 1
dx=du
then
(U+2)/U Du
= U/2 + 2/U
=2u^-1 + 2lnU
solve from there
12. (Original post by Notnek)
2+2ln2 - 1+2ln1

1+2ln2

These are the same thing since ln(1) = 0.

For future questions please post all your working at the start so that we know where you're stuck.
Oh, I see. Thank you very much . There is one other question which was similar:
13. (Original post by simon0)
Note: (a+b) - (c+d) = a + b - c - d, (note the bold emphasis).
What bold emphasis?
14. (Original post by MrToodles4)
What bold emphasis?
Sorry, it is the last minus sign.

What you have done looks fine except for the final part but it is fine in this case since ln(1)=0.

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