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# C4 Integration help please! watch

Integrate with respect to x:

x / ((x - 1)^0.5)

I attempted this:

(x)*((x - 1)^-0.5)

u = x -> du/dx = 1

dv/dx = ((x - 1) ^-0.5) -> v = 2*((x - 1)^0.5)

= 2x*((x - 1)^0.5) - Int 2*((x - 1)^0.5)

so the answer I got was ... 2x*((x - 1)^0.5) - (4/3)*((x - 1)^(3/2))

However the answer is: (2/3)(x + 2)((x - 1)^0.5)

Any help much appreciated!
2. (Original post by moonymeen)

Integrate with respect to x:

x / ((x - 1)^0.5)

I attempted this:

(x)*((x - 1)^-0.5)

u = x -> du/dx = 1

dv/dx = ((x - 1) ^-0.5) -> v = 2*((x - 1)^0.5)

= 2x*((x - 1)^0.5) - Int 2*((x - 1)^0.5)

so the answer I got was ... 2x*((x - 1)^0.5) - (4/3)*((x - 1)^(3/2))

However the answer is: (2/3)(x + 2)((x - 1)^0.5)

Any help much appreciated!
You don't need integration by parts for this. I would do one of the following:

1) Note that .

2) Use a substitution of .
3. (Original post by moonymeen)
so the answer I got was ... 2x*((x - 1)^0.5) - (4/3)*((x - 1)^(3/2))

However the answer is: (2/3)(x + 2)((x - 1)^0.5)

Any help much appreciated!
The answer you got is correct as far as you have gone; you need to take out a common factor of (x-1)^0.5 and rearrange the rest, and you'll then be able to take out a factor of 2/3 as well.
4. (Original post by Farhan.Hanif93)
You don't need integration by parts for this. I would do one of the following:

1) Note that .

2) Use a substitution of .
What the balls have you done in that first solution?
Edit: Even after you've changed it- how did you get from (x+1-1) to that next step?
5. (Original post by MartinKellyisagod)
What the balls have you done in that first solution?
Just remembered that and then expanded the brackets in a helpful way.

I had misread the OP at first though, I thought it was a bit odd that they would pose the question with a negative index in the denominator. Fixed it now.
6. (Original post by MartinKellyisagod)
Edit: Even after you've changed it- how did you get from (x+1-1) to that next step?
, which leads to what I wrote.
7. (Original post by Farhan.Hanif93)
, which leads to what I wrote.
Genius! Why's that first term = to (x-1)^(3/2) though? Am I being retarded? lol
8. (Original post by MartinKellyisagod)
Genius! Why's that first term = to (x-1)^(3/2) though? Am I being retarded? lol
Nope, I've edited my post since you first quoted me because originally I didn't read the OP's post correctly.
9. (Original post by Farhan.Hanif93)
You don't need integration by parts for this. I would do one of the following:

1) Note that .

2) Use a substitution of .
Wouldn't it be easier to bring the (x-1)^1/2 to the top and then use parts?
Wouldn't it be easier to bring the (x-1)^1/2 to the top and then use parts?
In my opinion, parts should only be used as a last resort. Mainly in cases where it's either obviously the only way or where you've tried to manipulate the integrand/use a substitution but have gotten nowhere. It's often more legwork than you need to put in.
11. (Original post by Farhan.Hanif93)
In my opinion, parts should only be used as a last resort. Mainly in cases where it's either obviously the only way or where you've tried to manipulate the integrand/use a substitution but have gotten nowhere. It's often more legwork than you need to put in.
Fair play.
12. (Original post by Farhan.Hanif93)
You don't need integration by parts for this. I would do one of the following:

1) Note that .
How did you possibly see that lol?
13. (Original post by Farhan.Hanif93)
You don't need integration by parts for this. I would do one of the following:

1) Note that .

2) Use a substitution of .

(Original post by ghostwalker)
The answer you got is correct as far as you have gone; you need to take out a common factor of (x-1)^0.5 and rearrange the rest, and you'll then be able to take out a factor of 2/3 as well.
I understand it now. Thank you very much for your time and help
14. (Original post by Thrug)
How did you possibly see that lol?
Lots of practice, understanding and basically this: http://en.wikipedia.org/wiki/Mathematical_maturity
15. (Original post by EEngWillow)
Lots of practice, understanding and basically this: http://en.wikipedia.org/wiki/Mathematical_maturity
I haven't completed my Further Maths A-Level yet; I certainly wouldn't describe myself as mathematically mature!
16. (Original post by Farhan.Hanif93)
I haven't completed my Further Maths A-Level yet; I certainly wouldn't describe myself as mathematically mature!
I'm aware of what level you're at

Maturity is something that develops across the entire time domain. It wouldn't be inaccurate to claim you're more mature mathematically than most A Level students, imo.

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