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Integration P3 watch

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    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?
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    ur integration of 2u/u+1 is wrong
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    (Original post by rockindemon)
    ur integration of 2u/u+1 is wrong
    What is it then ?
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    (Original post by imasillynarb)
    What is it then ?
    Its 2/(1+u)
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    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
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    (Original post by Silly Sally)
    Its 2/(1+u)
    Sorry talking nonsense
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    no silly sam it is not 2/*** it is 2u/***
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    lol. too quick
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    (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)
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    (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??
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    (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
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    (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? :confused:
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    (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)
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    2u/(1+u)

    split into partial fractions to give:

    2 - 2/(1+u)

    integrate to give 2u - 2ln(1+u)
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    (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
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    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)
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    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

    = the correct answer.
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    (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)
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    (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)
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    At last, Im gonna have a break now
 
 
 
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