Integrating $\int \frac{x^2 - 1}{x^4 + 1}dx$ - Help Needed

[VietDao29]

Hi,
Can you guys just throw me a hint on how to integrate:
\int \frac{x^2 - 1}{x^4 + 1}dx
I am completely lost. I don't even have any idea how to start.
Any help will be appreciated.
Thanks.
Viet Dao,

 

[Jameson]

Hmmm... stupid plus sign!

I did this on an online integrator and got the difference of two logs, which would make me think partial fractions, although in this case that would deal with complex numbers.

 

[Pyrrhus]

You can solve this with Partial fractions.

x^4 + 1 = (x^2 - \sqrt{2}x +1) (x^2 + \sqrt{2}x +1)

 

[Jameson]

That's it. The answer online used that same form.

 

[HallsofIvy]

To see how Cyclovenom got that, if you let y= x², x⁴+ 1 becomes y²+ 1. The roots to y²+ 1= 0 are y= +/- i. That means we have y= x²= i so that x= \frac{\sqrt{2}}{2}(1+i) and x= -\frac{\sqrt{2}}{2}(1+i). We also have y= x²= -i so that
x= \frac{\sqrt{2}}{2}(1- i) and x= -\frac{\sqrt{2}}{2}(1- i), the four complex roots of x⁴+ 1= 0. That tells us that x⁴+ 1 factors as (x-\sqrt{2}}{2}(1+i))(x+\sqrt{2}}{2}(1+i))(x-\frac{\sqrt{2}}{2}(1- i))(x+\frac{\sqrt{2}}{2}(1- i)).
We can rearrange the factors as (x-\sqrt{2}}{2}(1+i))(x-\frac{\sqrt{2}}{2}(1- i))(x+\sqrt{2}}{2}(1+i))))(x+\frac{\sqrt{2}}{2}(1- i)) to get the two real factors Cyclovenom gives.

 

[jdavel]

To see how Cyclovenom got that, if you let y= x², x⁴+ 1 becomes y²+ 1. The roots to y²+ 1= 0 are y= +/- i. That means we have y= x²= i so that x= \frac{\sqrt{2}}{2}(1+i) and x= -\frac{\sqrt{2}}{2}(1+i). We also have y= x²= -i so that
x= \frac{\sqrt{2}}{2}(1- i) and x= -\frac{\sqrt{2}}{2}(1- i), the four complex roots of x⁴+ 1= 0. That tells us that x⁴+ 1 factors as (x-\sqrt{2}}{2}(1+i))(x+\sqrt{2}}{2}(1+i))(x-\frac{\sqrt{2}}{2}(1- i))(x+\frac{\sqrt{2}}{2}(1- i)).
We can rearrange the factors as (x-\sqrt{2}}{2}(1+i))(x-\frac{\sqrt{2}}{2}(1- i))(x+\sqrt{2}}{2}(1+i))))(x+\frac{\sqrt{2}}{2}(1- i)) to get the two real factors Cyclovenom gives.

Or you can just say that

x^4 + 1 = (x^2 +/- ax +/- 1)(x^2 -/+ ax +/- 1)

and in a few tries you get that a = sqrt(2).

Trial and error isn't elegant, but in this case it's a lot faster.

 

[dextercioby]

How about that

x^{4}+1=x^{4}+2x^{2}+1-2x^{2}=\left(x^{2}+1\right)^{2}-\left(\sqrt{2}x\right)^{2}

and then use the square difference formula...?

Daniel.

 

[jdavel]

How about that

x^{4}+1=x^{4}+2x^{2}+1-2x^{2}=\left(x^{2}+1\right)^{2}-\left(\sqrt{2}x\right)^{2}

and then use the square difference formula...?

Daniel.

Daniel,

Very elegant and very fast. As usual, you win!

jdl

 

[GCT]

To see how Cyclovenom got that, if you let y= x2, x4+ 1 becomes y2+ 1. The roots to y2+ 1= 0 are y= +/- i. That means we have y= x2= i so that and . We also have y= x2= -i so that
and , the four complex roots of x4+ 1= 0. That tells us that x4+ 1 factors as .
We can rearrange the factors as to get the two real factors Cyclovenom gives.
__________________
"Euclid alone has looked on beauty bare"

I'm not quite sure what you did here. Can you or anyone else here point out where I might learn the details of this? References, such as texts, online sites, the theorems, title of the subject, or even a brief explanation.

 

[VietDao29]

I get it now. It's not as 'hard' as I used to think.
Thanks very much,
Viet Dao,

 

[SphericalStrife]

could you use u-substition to find the answer too?

 

[Nylex]

could you use u-substition to find the answer too?

Nope, at least it doesn't look like it.

 

[bblueblob]

If you take the next two numbers in this sequence and multiply them together, what number will you get?

594, 487, 566, 493, 310, 447, ____, ____

 

[Jameson]

Good try. You've already made a threadt for this topic. Don't double post. It makes people mad here and it's against the rules.