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Math Help - 3 integration technique problems

  1. #1
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    3 integration technique problems

    the first is a division problem,

    int of 3x^4 + 3x^3 - 5x^2 + x - 1 / x^2 + x - 2 dx
    3x + 0 - 11
    ________________________
    i started by dividing it and getting x^2 + x - 2 | 3x^4 + 3x^3 - 5x^2 x - 1
    -(3x^4 + 3x^3 - 6x^2)
    - 11x^2 + x - 1
    -(-11x^2 + 11x +22)
    12x + 23


    i dont exactly remember long division, but thats where im a so far, do i now sub that 3x-11 thing back into the integral? what about the remainder

    the second problem is
    int x^3 + 3x^2 + x - 9/ (x^2 + 1)(x^2 + 3) dx

    is this decomposing fractions? would it be x^3 + 3x^2 + x - 9= Ax+b/(x^2 + 3) and cx + d/(x^2 + 1)
    im completely lost on this guy

    the third problem, i thing i made the most progress on

    int dx/ sqrt x - 4throot x

    i made u= x^ 1/4
    then u ^4= x
    then 4u^3 du = dx

    sub it all back in
    int 4u^3 du/ (u^2) ^1/4 - (u^4)^1/2

    i dont know where to go from here.

    I really could use some help with these. Im not looking for an easy answer, just set me on the right path and ill be grateful
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  2. #2
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    Quote Originally Posted by gbux512 View Post
    the third problem, i thing i made the most progress on

    int dx/ sqrt x - 4throot x

    i made u= x^ 1/4
    \displaystyle \int \frac{dx}{\sqrt{x}- \sqrt[4]{x}}

    make u = \sqrt[4]{x} \implies du = \frac{1}{4x^{\frac{3}{4}}}

    Giving \displaystyle 4\int \frac{u^3}{u^2-u}~du = 4\int \frac{u^2}{u-1}~du

    Then use long division, after this integrate term by term.
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