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# C4 maths integration edexcel, exercie 6F question watch

1. Hi there, if anyone can explain the process of integrating question 3c on exercise 6F.

It is integrate with respect to x, 1/1+(x-1)^1/2 with upper limit 5 and lower limit 2, with the substitution u^2=x-1

Apologies for the poor typing

I looked at the solution and it goes another path from what I got.
In the end, i got integrate with limits 2 and 1 2du/u+1 but the solutions goes to do 2-2/u+1
I understand what they did to get the 2-2/u+1 but why did they do that instead of just integrating my answer?

I know how i typed it is poor, but if any of you have th c4 edexcel book, it is exercise 6F question 3c
2. (Original post by THESTRESS)
It is integrate with respect to x, 1/1+(x-1)^1/2 with upper limit 5 and lower limit 2, with the substitution u^2=x-1

In the end, i got integrate with limits 2 and 1 2du/u+1 but the solutions goes to do 2-2/u+1
I understand what they did to get the 2-2/u+1 but why did they do that instead of just integrating my answer?
c
Should your integral be 2u/(u+1) ? If so it counts as top heavy, so you need to divide (or the equivalent) before integrating.
3. (Original post by tiny hobbit)
Should your integral be 2u/(u+1) ? If so it counts as top heavy, so you need to divide (or the equivalent) before integrating.
Hi there,

if you look at this working out, it is the steps indicated with the arrow which i dont understand the reason behind.
why could i just not integrate the step above the marked with the arrow?

thanks for the help

update: just realised what you said....and yes it is top heavy. so i divided it to get 2-1/(u+1) which still not the same??
Attached Images

4. (Original post by THESTRESS)
Hi there,

if you look at this working out, it is the steps indicated with the arrow which i dont understand the reason behind.
why could i just not integrate the step above the marked with the arrow?

thanks for the help

update: just realised what you said....and yes it is top heavy. so i divided it to get 2-1/(u+1) which still not the same??
When you "just" integrate 2u/(u+1) do you get 2ln(u+1) ? if so, try differentiating that and you won't get 2u/(u+1). That's why you need to divide first.

Try dividing again - it should come to 2 - 2/(u+1).
5. (Original post by tiny hobbit)
When you "just" integrate 2u/(u+1) do you get 2ln(u+1) ? if so, try differentiating that and you won't get 2u/(u+1). That's why you need to divide first.

Try dividing again - it should come to 2 - 2/(u+1).

Thank you very much, i understand and got the answer now.

but can i ask, in my first attempt to integrate what the first arrow is point to, i though i would minus by making the denominators the same, which got me 2u/u+1 again. Why can i not do this?

Also, could you just clarify that if i differentiate of 2ln(u+1) , i get 2/u+1 right?
Thanks
6. (Original post by THESTRESS)
but can i ask, in my first attempt to integrate what the first arrow is point to, i though i would minus by making the denominators the same, which got me 2u/u+1 again. Why can i not do this?
This just takes you back to where you came from. You need the split up version for integrating.

(Original post by THESTRESS)
Also, could you just clarify that if i differentiate of 2ln(u+1) , i get 2/u+1 right?
Thanks
Yes. That's why you need to split the original fraction up, as you know that this tells you that integrating 2/(u+1) gives 2ln(u+1)
7. hi, i know its a while later, but in my revision i stumbled on a question after this, which is xercise 6f question 3d.
in the marked place, why do i not divide here when it is top heavy?

8. (Original post by THESTRESS)
hi, i know its a while later, but in my revision i stumbled on a question after this, which is xercise 6f question 3d.
in the marked place, why do i not divide here when it is top heavy?

What's the question...?
9. (Original post by RDKGames)
What's the question...?
sorry, the photo did not attach.
The question is why we do not need to divide in this question where marked with an arrow, despite being improper whereas in the above chat i had to divide through?
Thanks
Attached Images

10. (Original post by THESTRESS)
sorry, the photo did not attach.
The question is why we do not need to divide in this question where marked with an arrow, despite being improper whereas in the above chat i had to divide through?
Thanks
They HAVE divided through. Since the bottom is just a single letter however, you can just put each bit of the top over u and cancel each fraction:

2u/u becomes 2, which integrates to give 2u.

2/u stays like that and integrates to give 2ln u.
11. (Original post by tiny hobbit)
They HAVE divided through. Since the bottom is just a single letter however, you can just put each bit of the top over u and cancel each fraction:

2u/u becomes 2, which integrates to give 2u.

2/u stays like that and integrates to give 2ln u.
Ohhh i see...thank you very much!

So just to clarify, in the example before the one i showed now, since the bottom is not just a single letter, but a letter + something else, it is better to do long division to simplify it?

Thank you so much!
12. (Original post by THESTRESS)
Ohhh i see...thank you very much!

So just to clarify, in the example before the one i showed now, since the bottom is not just a single letter, but a letter + something else, it is better to do long division to simplify it?

Thank you so much!
Yes if it is top-heavy, i.e. yes if the highest power on the top is equal to or higher than the highest power on the bottom.

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