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Math Help - Integrals, #1

  1. #1
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    Post Integrals, #1

    Hi
    Can you solve the problem no. 216 in the atachment ?!
    i know I will need the substitution u = ln t at the first

    and then ??
    I wasted 2 hours ;(
    then i tried k = u-1 or k = u+1 but it didnt work ..

    another Question: in problem no.220 :
    a = 1 , Right ?
    just check it if its right or not, You are not need to solve it for me

    Thank you
    sorry for my bad english ;x
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    Last edited by TWiX; June 8th 2009 at 05:47 AM. Reason: k=u+1 and k=u-1
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  2. #2
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    I got it !!

    it will = sec^-1 [ln t + 1] + C


    Solving by :
    Sqrt( u . (u+2) )
    = sqrt [ (u+0) . (u+1+1) ]
    = sqrt [ (u+1-1) . (u+1+1) ]
    then i will mulitplyed the brackets
    it will be = sqrt [ (u+1)^2 - 1 ]
    put k = u+1
    dk = du
    it will be k . Sqrt(k^2 -1)
    Trigonometric substitution k = sec θ

    ... etc

    Right ??
    One question : it allowed to multiply the brackets in BLUE making u+1 as an one thing??
    i hope u will understand what i mean !
    Again: Sorry for bad english
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  3. #3
    MHF Contributor Amer's Avatar
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    your integral

    \int \frac{dt}{t(1+ln(t))\sqrt{ln(t)(2+ln(t)}}

    let u = lnt .... du = dt/t .... tdu=dt

    \int \frac{({\color{red}\rlap{/}}t)du}{{\color{red}\rlap{/}}t(1+u)\sqrt{u(2+u)}}

    \int \frac{du}{(1+u)\sqrt{u^2+2u+1-1}}

    \int frac {du}{(1+u)\sqrt{(u+1)^2 -1 } }

    let x = u+1 ........dx = du

    \int \frac{dx}{x\sqrt{x^2 -1 }}

    I think you can solve this right
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  4. #4
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    thank you Amer,
    i got it
    funny Question!

    Check a in problem 220 plz
    it will be a = 1 right ??
    dont solve it for me
    just check it !

    Thank you ^^
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  5. #5
    MHF Contributor Amer's Avatar
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    a=1 that's correct
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  6. #6
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    Thank you.
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