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Math Help - Stick on evaluatin integrals, Please help

  1. #1
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    Stick on evaluatin integrals, Please help

    Question :
    integral from 1 to 2 of (4+u^2)/u^3

    My work:

    integreal 4/u^3+ integral u^2/u^3

    4integral 1/u^3 + integral (u^2)(1/u^3)

    4 integral u^-3 + integral (u^2)(u^-3)

    4 integral (u^-2)/-2 + integral (u^3/3)(u^-2/-2)

    4 [[(2^-2)/-2] -[(1^-2)/-2]} +{ [(2^3/3)(2^-2/-2)]-[(1^3/3)(1^-2/-2)]}

    Now im stuck because i want to make my answer clean in a fraction. i can solve it usiing decimals but how would i re write it soo i end up dealing with fractions?
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  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
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    u^2/u^3=1/u
    =================

    (4+u^2)/u^3=4/u^3+1/u

    int(4/u^3+1/u)=int(4/u^3)+int(1/u)=int(4u^-3)+ln(u)+C=(4u^-2)/(-2)+ln(u)+C=-2(1/u^2)+ln(u)+C
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  3. #3
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    Quote Originally Posted by Also sprach Zarathustra View Post
    u^2/u^3=1/u
    =================

    (4+u^2)/u^3=4/u^3+1/u

    int(4/u^3+1/u)=int(4/u^3)+int(1/u)=int(4u^-3)+ln(u)+C=(4u^-2)/(-2)+ln(u)+C=-2(1/u^2)+ln(u)+C
    subbing in [1,2] into -2(1/u^2)+ln(u) i got
    =[[-2(1/2^2)]-[-2(1/1^2)]] +[ln(2)- ln(1)]
    which i then got 1.5+ln2 which is 2.19?

    It looks really ugly, did i do something wrong?
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  4. #4
    MHF Contributor Also sprach Zarathustra's Avatar
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    Yes! This is the correct answer!
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