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Math Help - [SOLVED] Integral

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    [SOLVED] Integral

    \int\frac{dy}{1+y^4}

    \int\frac{dy}{1+(y^2)^2}

    Does this fit the arctan integral or not since the y term is squared?
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    Quote Originally Posted by dwsmith View Post
    \int\frac{dy}{1+y^4}

    \int\frac{dy}{1+(y^2)^2}

    Does this fit the arctan integral or not since the y term is squared?
    if I remember correctly ...

    \frac{1}{y^4+1} = \frac{1}{(y^2 + \sqrt{2} \, y + 1)(y^2 - \sqrt{2} \, y + 1)}

    partial fractions, I do believe.
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    Quote Originally Posted by dwsmith View Post
    \int\frac{dy}{1+y^4}

    \int\frac{dy}{1+(y^2)^2}

    Does this fit the arctan integral or not since the y term is squared?
    Skeeter is correct and this should be done via partial fractions. I do believe this does not fit the form for arctan because when we do chain rule we will end up with an extra 2y on top.
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    Quote Originally Posted by AllanCuz View Post
    Skeeter is correct and this should be done via partial fractions. I do believe this does not fit the form for arctan because when we do chain rule we will end up with an extra 2y on top.
    I understand that; that is why I asked the question.
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  5. #5
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by dwsmith View Post
    I understand that; that is why I asked the question.
    I thought you were asking if it would fit and I said no, that's my final answer lol.

    Skeeter makes it harder then need be though (I think...).

    For partial fractions,

     \frac{dy}{1+(y^2)^2 }

    Let P=y^2 and dp=2ydy

    Noting that  \sqrt{P}=y

    We then get,

     \frac{1}{2 \sqrt{P} } \frac{dp}{1+p^2} = \frac{1}{2} \frac{ dp}{ \sqrt{P} ( 1 + p^2 ) }

    I believe this will make it easiar to solve.
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  6. #6
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    Quote Originally Posted by skeeter View Post
    if I remember correctly ...

    \frac{1}{y^4+1} = \frac{1}{(y^2 + \sqrt{2} \, y + 1)(y^2 -  \sqrt{2} \, y + 1)}

    partial fractions, I do believe.
    The coefficient matrix I obtained was:

    \begin{bmatrix}<br />
1 & 0 & 1 & 0 & :0\\ <br />
-\sqrt{2} & 1 & \sqrt{2} & 1 & :0\\ <br />
1 & \sqrt{2} & 1 & \sqrt{2} & :0\\ <br />
0 & 1 & 0 & 1 & :1<br />
\end{bmatrix} which is inconsistent.

    Quote Originally Posted by AllanCuz View Post
     \frac{1}{2 \sqrt{P} } \frac{dp}{1+p^2} = \frac{1}{2} \frac{ dp}{ \sqrt{P} ( 1 + p^2 ) }

    I believe this will make it easiar to solve.
    And for yours, I obtained:

    \begin{bmatrix}<br />
1 & 0 & 0 & :1\\ <br />
1 & 0 & 0 & :0\\ <br />
0 & 1 & 0 & :0\\ <br />
0 & 0 & 1 & :0 <br />
\end{bmatrix} which is also inconsistent.
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  7. #7
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    \begin{bmatrix}<br />
1 & 0 & 1 & 0 & :0\\ <br />
-\sqrt{2} & 1 & \sqrt{2} & 1 & :0\\ <br />
1 & \textcolor{red}{-}\sqrt{2} & 1 & \sqrt{2} & :0\\ <br />
0 & 1 & 0 & 1 & :1<br />
\end{bmatrix}

    A = \frac{1}{2\sqrt{2}}

    B = \frac{1}{2}

    C = -\frac{1}{2\sqrt{2}}

    D = \frac{1}{2}
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