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Math Help - Integration

  1. #1
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    Integration

    I need a bit of help integrating

    X^2/(X^2-4X+5)
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  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
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    Quote Originally Posted by qwerty10 View Post
    I need a bit of help integrating

    X^2/(X^2-4X+5)

    Hint:

    \frac{x^2}{x^2-4x+5}=\frac{4x-8}{x^2-4x+5}+\frac{3}{x^2-4x+5}+1
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  3. #3
    MHF Contributor Unknown008's Avatar
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    1. Do a long division.
    2. Apply:

    \int \dfrac{f'(x)}{f(x)} dx = ln |f(x)| + C
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  4. #4
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    Thank you
    Is the correct way to continue by

    let u=x^2-4x+5
    du=(2x-4)dx

    so (x-2)dx=du/4

    After this I am not quite sure what to do
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  5. #5
    MHF Contributor Unknown008's Avatar
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    \int \frac{x^2}{x^2-4x+5} dx = \int 1+ \frac{4x-5}{x^2-4x+5} dx = \int 1 dx + \int \frac{4x-5}{x^2-4x+5} dx

     = \int 1 dx + \int \frac{4x-8}{x^2-4x+5} dx + \int \frac{3}{x^2-4x+5} dx

     =x + 2ln |x^2-4x+5| + \int \frac{3}{x^2-4x+5} dx

    Can you continue now?
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  6. #6
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by qwerty10 View Post
    Thank you
    Is the correct way to continue by

    let u=x^2-4x+5
    du=(2x-4)dx

    so (x-2)dx=du/4 Not Needed rather dx = du/(2x-4)

    After this I am not quite sure what to do
    Look at the reply by Also sprach Zarathustra

    \int  \frac{x^2}{x^2-4x+5}dx= \int \frac{4x-8}{x^2-4x+5}dx+\int \frac{3}{x^2-4x+5}dx + \int 1 dx

    use the substitution rule for the first term only:

     \int \frac{4x-8}{x^2-4x+5}dx = \int \frac{2(x-4)}{x^2-4x+5}

    sub. rule gives you:

    \int \frac{2}{u} du . complete it..


    For the second term:  \int \frac{3}{x^2-4x+5}dx = \int \frac{3}{1+(x-2)^{2}} dx

    you should be able to finish it off now..
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  7. #7
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    Thank you

    I have worked the answer out to be 2ln(x^2-4x+5) +3tan^-1(x-2) + x+C ??
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  8. #8
    MHF Contributor harish21's Avatar
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