[SOLVED] Integral

dwsmith

MHF Hall of Honor
Mar 2010
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Florida
\(\displaystyle \int\frac{dy}{1+y^4}\)

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

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

skeeter

MHF Helper
Jun 2008
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North Texas
\(\displaystyle \int\frac{dy}{1+y^4}\)

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

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.
 
Apr 2010
384
153
Canada
\(\displaystyle \int\frac{dy}{1+y^4}\)

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

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.
 

dwsmith

MHF Hall of Honor
Mar 2010
3,093
582
Florida
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.
 
Apr 2010
384
153
Canada
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.
 

dwsmith

MHF Hall of Honor
Mar 2010
3,093
582
Florida
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.

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

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.
 

skeeter

MHF Helper
Jun 2008
16,216
6,764
North Texas
\(\displaystyle \begin{bmatrix}
1 & 0 & 1 & 0 & :0\\
-\sqrt{2} & 1 & \sqrt{2} & 1 & :0\\
1 & \textcolor{red}{-}\sqrt{2} & 1 & \sqrt{2} & :0\\
0 & 1 & 0 & 1 & :1
\end{bmatrix}\)

\(\displaystyle A = \frac{1}{2\sqrt{2}}\)

\(\displaystyle B = \frac{1}{2}\)

\(\displaystyle C = -\frac{1}{2\sqrt{2}}\)

\(\displaystyle D = \frac{1}{2}\)
 
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