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Math Help - another integration problem

  1. #1
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    another integration problem

    (x-2)/(x^2+x^4) dy
    I tried this way, but the result didn't seem right.
    (AX+B)(X^2+1)+(C)/(X)+(D)/(X^2)
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  2. #2
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    Quote Originally Posted by questionboy View Post
    (x-2)/(x^2+x^4) dy
    I tried this way, but the result didn't seem right.
    (AX+B)(X^2+1)+(C)/(X)+(D)/(X^2)
    Your partial fraction decomposition is correct. Continue using it.
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  3. #3
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    really?
    then i have
    (-x+2)/(x^2+1)+1/x+(-2/x^2) dx

    then
    {ln|x|-2ln|x^2|} +{(-x+2)/(x^2+1) dx}
    Im stuck here

    i tried this
    {(-x+2)/(x^2+1) dx}
    a=1
    x=tany
    dx=secy^2 dy
    [(-tany+2)*secy^2]/ (secy^2)
    =-tany+2
    i don't know what to do now
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  4. #4
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    Quote Originally Posted by questionboy View Post
    really?
    then i have
    (-x+2)/(x^2+1)+1/x+(-2/x^2) dx

    then
    {ln|x|-2ln|x^2|} +{(-x+2)/(x^2+1) dx}
    Im stuck here

    i tried this
    {(-x+2)/(x^2+1) dx}
    a=1
    x=tany
    dx=secy^2 dy
    [(-tany+2)*secy^2]/ (secy^2)
    =-tany+2
    i don't know what to do now
    All integrals are standard forms. Omitting the C's:

    \int \frac{1}{x^2} \, dx = \int x^{-2} \, dx = - \frac{1}{x}.

    \int \frac{-x + 2}{x^2 + 1} = - \int \frac{x}{x^2 + 1} \, dx + \int \frac{2}{x^2 + 1} \, dx = - \frac{1}{2} \ln |x^2 + 2| + 2 \tan^{-1} x.
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  5. #5
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    Thank you very much.Master
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