$\displaystyle \int\frac{dy}{1+y^4}$
$\displaystyle \int\frac{dy}{1+(y^2)^2}$
Does this fit the arctan integral or not since the y term is squared?
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,
$\displaystyle \frac{dy}{1+(y^2)^2 } $
Let $\displaystyle P=y^2$ and $\displaystyle dp=2ydy$
Noting that $\displaystyle \sqrt{P}=y$
We then get,
$\displaystyle \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.
The coefficient matrix I obtained was:
$\displaystyle \begin{bmatrix}
1 & 0 & 1 & 0 & :0\\
-\sqrt{2} & 1 & \sqrt{2} & 1 & :0\\
1 & \sqrt{2} & 1 & \sqrt{2} & :0\\
0 & 1 & 0 & 1 & :1
\end{bmatrix}$ which is inconsistent.
And for yours, I obtained:
$\displaystyle \begin{bmatrix}
1 & 0 & 0 & :1\\
1 & 0 & 0 & :0\\
0 & 1 & 0 & :0\\
0 & 0 & 1 & :0
\end{bmatrix}$ which is also inconsistent.
$\displaystyle \begin{bmatrix}
1 & 0 & 1 & 0 & :0\\
-\sqrt{2} & 1 & \sqrt{2} & 1 & :0\\
1 & \textcolor{red}{-}\sqrt{2} & 1 & \sqrt{2} & :0\\
0 & 1 & 0 & 1 & :1
\end{bmatrix}$
$\displaystyle A = \frac{1}{2\sqrt{2}}$
$\displaystyle B = \frac{1}{2}$
$\displaystyle C = -\frac{1}{2\sqrt{2}}$
$\displaystyle D = \frac{1}{2}$