# [SOLVED] Integral

#### dwsmith

MHF Hall of Honor
$$\displaystyle \int\frac{dy}{1+y^4}$$

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

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

#### skeeter

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

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

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.

#### AllanCuz

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

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

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
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.

#### AllanCuz

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

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

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

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

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

• dwsmith