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Math Help - PARTS HELP IM NEW HERE

  1. #1
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    PARTS HELP IM NEW HERE

    i have a problem INTEGRAL [1/(X^3-X^2)] CAN SOME ONE HELP ME I GOT AS FAR AS [(A/X)+(B/X^2)+(C/X-1)] BUT I CANT SEEM TO GET THEM TO CANCEL AND GET RID OF THE DENOMINATORS LIKE I HAVE IN PREVIOUS PROBLEMS ANY HINTS OR SOLUTIONS
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by NEWBIE View Post
    i have a problem INTEGRAL [1/(X^3-X^2)] CAN SOME ONE HELP ME I GOT AS FAR AS [(A/X)+(B/X^2)+(C/X-1)] BUT I CANT SEEM TO GET THEM TO CANCEL AND GET RID OF THE DENOMINATORS LIKE I HAVE IN PREVIOUS PROBLEMS ANY HINTS OR SOLUTIONS
    stop shouting.

    note that you have:

    \frac 1{x^2(x - 1)} = \frac Ax + \frac B{x^2} + \frac C{x - 1}

    multiply both sides by the LCD, we get:

    1 = Ax(x - 1) + B(x - 1) + Cx^2 ....................(1)

    let x = 0 in (1) and solve for B

    then let x = 1 in (1) and solve for C

    then let x = anything but 0 and 1 in (1) and solve for A (note that you will have to plug in the values you found for B and C
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  3. #3
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    \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x-1} = \frac{1}{x^2(x-1)}

    Ax(x-1) + B(x-1) + Cx^2 = 1

    x = 1 ... C = 1

    x = 0 ... B = -1

    x = 2 ... 2A + B + 4C = 1 ... A = -1

    -\frac{1}{x} - \frac{1}{x^2} + \frac{1}{x-1} = \frac{1}{x^2(x-1)}

    now integrate
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  4. #4
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    <br /> <br />
    " alt="1 = Ax(x - 1) + B(x - 1) + Cx^2
    " />

    i got that but why did you choose 0


    i did

    <br /> <br />
(a+c)x^2 + (-a+b) - b

    " alt="(a+c)x^2 + (-a+b) - b

    " />
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  5. #5
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    1 = Ax(x - 1) + B(x - 1) + Cx^2


    i got that but why did you choose 0


    i did



    (a+c)x^2 + (-a+b)x - b


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  6. #6
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    -\frac{1}{x} - \frac{1}{x^2} + \frac{1}{x-1} = \frac{1}{x^2(x-1)}


    what would be the first step in integrating this
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  7. #7
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by NEWBIE View Post


    1 = Ax(x - 1) + B(x - 1) + Cx^2


    i got that but why did you choose 0


    i did



    (a+c)x^2 + (-a+b)x - b


    we chose zero to wipe out A and C and leave B. we chose 1 to wipe out A and B and leave C
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  8. #8
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by NEWBIE View Post
    -\frac{1}{x} - \frac{1}{x^2} + \frac{1}{x-1} = \frac{1}{x^2(x-1)}


    what would be the first step in integrating this
    do you notice that your code is not working?

    please stop using color text if you don't know how. it is messing with the format
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  9. #9
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    sorry i tried copying the code that was used
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  10. #10
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    (a+c)x^2 + (-a+b)x - b


    i set -b=1

    -a+b=1

    -2+c=1\


    where did i go wrong





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  11. #11
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    hello
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  12. #12
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by NEWBIE View Post
    hello
    firstly, bumping is against the rules

    secondly, you seem to ignore everything else we told you
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  13. #13
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    Alternatively, you can pull out x^2 common factor and substitute u = \frac{1}{x}. You should get it in the form of \int \frac{u}{1-u}~du which is much easier to solve.
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  14. #14
    Super Member Showcase_22's Avatar
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    lol, you're quite a character!

    Here's what I would do:

    \int\frac{1}{x^3-x^2}dx

    \int\frac{1}{x^2(x-1)}dx

    \int\frac{1+x-x}{x^2(x-1)}dx

    -\int\frac{x-1-x}{x^2(x-1)}dx

    -\int\frac{x-1}{x^2(x-1)}-\frac{x}{x^2(x-1)}dx

    -\int\frac{1}{x^2}-\frac{1}{x(x-1)}dx

    -\int\frac{1}{x^2}+\frac{x-1-x}{x(x-1)}dx

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

    You can choose to multiply it by -1 if you like so you don't have a minus integral.

    There you go NEWBIE, now stop bumping!!!
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