1. ## [SOLVED] Integral

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

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

Does this fit the arctan integral or not since the y term is squared?

2. Originally Posted by dwsmith
$\displaystyle \int\frac{dy}{1+y^4}$

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

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

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

partial fractions, I do believe.

3. Originally Posted by dwsmith
$\displaystyle \int\frac{dy}{1+y^4}$

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

Does this fit the arctan integral or not since the y term is squared?
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.

4. Originally Posted by AllanCuz
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.

5. Originally Posted by dwsmith
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,

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

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.

6. Originally Posted by skeeter
if I remember correctly ...

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

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

$\displaystyle \begin{bmatrix} 1 & 0 & 1 & 0 & :0\\ -\sqrt{2} & 1 & \sqrt{2} & 1 & :0\\ 1 & \sqrt{2} & 1 & \sqrt{2} & :0\\ 0 & 1 & 0 & 1 & :1 \end{bmatrix}$ which is inconsistent.

Originally Posted by AllanCuz
$\displaystyle \frac{1}{2 \sqrt{P} } \frac{dp}{1+p^2} = \frac{1}{2} \frac{ dp}{ \sqrt{P} ( 1 + p^2 ) }$

I believe this will make it easiar to solve.
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.

7. $\displaystyle \begin{bmatrix} 1 & 0 & 1 & 0 & :0\\ -\sqrt{2} & 1 & \sqrt{2} & 1 & :0\\ 1 & \textcolor{red}{-}\sqrt{2} & 1 & \sqrt{2} & :0\\ 0 & 1 & 0 & 1 & :1 \end{bmatrix}$

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

$\displaystyle B = \frac{1}{2}$

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

$\displaystyle D = \frac{1}{2}$