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# Integration P3 watch

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1. Using the substituion u^2 = x-1 prove that

Integral(between 5 and 2) 1/[1+ root(x-1)] = 2-ln(9/4)

What I did:

2u.du/dx = 1

2u.du = dx

u = root(x-1)

1/[1+u] x dx

= 2u/[1+u] du

Integrating that gives:

2uln(1+u)

Change the bounds gives 2 and 1

= ln(81/4)

Where have I gone wrong?
2. ur integration of 2u/u+1 is wrong
3. (Original post by rockindemon)
ur integration of 2u/u+1 is wrong
What is it then ?
4. (Original post by imasillynarb)
What is it then ?
Its 2/(1+u)
5. use by parts

u = 2u
du = 2

dv= (1+u)-1
v=ln(1+u)

integrate gives

2uln(1+u) - 2/1+u

put x's bk in then add limits or apply limits to u first gives the answer required
6. (Original post by Silly Sally)
Its 2/(1+u)
Sorry talking nonsense
7. no silly sam it is not 2/*** it is 2u/***
8. lol. too quick
9. (Original post by rockindemon)
use by parts

u = 2u
du = 2

dv= (1+u)-1
v=ln(1+u)

integrate gives

2uln(1+u) - 2/1+u

put x's bk in then add limits or apply limits to u first gives the answer required
Or you could use partial fractions to split up 2u/(1+u)
10. (Original post by rockindemon)
use by parts

u = 2u
du = 2

dv= (1+u)-1
v=ln(1+u)

integrate gives

2uln(1+u) - 2/1+u

put x's bk in then add limits or apply limits to u first gives the answer required
I get 2uln(1+u) - Integral of 2ln(1+u) which is unintegratable??
11. (Original post by Silly Sally)
Or you could use partial fractions to split up 2u/(1+u)
This time i am not talking nonsense - split using partial fractiona and integrate
12. (Original post by rockindemon)
use by parts

u = 2u
du = 2

dv= (1+u)-1
v=ln(1+u)

integrate gives

2uln(1+u) - 2/1+u

put x's bk in then add limits or apply limits to u first gives the answer required
hold on a mo... uv - int: u.dv
i see where you got 2u ln(1+u) but i dont think the integral of 2ln(1+u) is 2/1+u.. or am I doing something wrong?
13. (Original post by Silly Sally)
This time i am not talking nonsense - split using partial fractiona and integrate
yeah that sounds alot easier than what i'm trying (unsuccessfully)
14. 2u/(1+u)

split into partial fractions to give:

2 - 2/(1+u)

integrate to give 2u - 2ln(1+u)
15. (Original post by imasillynarb)
I get 2uln(1+u) - Integral of 2ln(1+u) which is unintegratable??
hehe... unintegratable..

you can integrate it, but you have to do it by parts .. again!
for the integral of ln x, imagine its 1 * ln x, and let v = lnx and du = 1
16. Im still getting the wrong answer

Using partial fractions:

2 - 2/(1 + u)

= [2u - 2ln(1+u)] between 2 and 1

= [4 - 2ln3] - [2 - 2ln2]

= 4 - 2 -ln9 + ln4
=2 - ln(4/9)
17. Imasillynarb, u were right, it becomes
Int (2u/(1+u)) du

= ((2u + 2-2)/(1+u) du = 2(u+1)/(u+1) - 2/1+u du = 2 - 2/1+u du

[2u - 2ln|1+u|] between 2 and 1

18. (Original post by imasillynarb)
Im still getting the wrong answer

Using partial fractions:

2 - 2/(1 + u)

= [2u - 2ln(1+u)] between 2 and 1

= [4 - 2ln3] - [2 - 2ln2]

= 4 - 2 -ln9 + ln4
=2 - ln(4/9)
i got the same i think we're integrating the wrong thing.. oh wait a minute, ln 4-ln9 is + ln(4/9) which is the same as -ln(9/4)
19. (Original post by imasillynarb)
Im still getting the wrong answer

Using partial fractions:

2 - 2/(1 + u)

= [2u - 2ln(1+u)] between 2 and 1

= [4 - 2ln3] - [2 - 2ln2]

= 4 - 2 -ln9 + ln4
=2 - ln(4/9)
Nearly there, you have just made a little mistake. On the line

4 - 2 - ln9 + ln4
then:

2 - (ln9 - ln4)
2 - ln(9/4)
20. At last, Im gonna have a break now

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