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

• VietDao29
Good luck with your problem.In summary, the conversation discusses various methods and strategies for solving the given integral. The most efficient method is using the square difference formula, while other options such as partial fractions and u-substitution are also mentioned. The conversation also briefly touches on the topic of finding the next numbers in a sequence.
VietDao29
Homework Helper
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,

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.

You can solve this with Partial fractions.

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

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

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.

HallsofIvy said:
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.

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

and then use the square difference formula...?

Daniel.

dextercioby said:

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

and then use the square difference formula...?

Daniel.

Daniel,

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

jdl

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.

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

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

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

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

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, ____, ____

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.

## 1. What is the general process for integrating a rational function?

The general process for integrating a rational function is to first rewrite the function in the form of partial fractions, if possible. Then, use the power rule for integration to integrate each term separately. Finally, add any necessary constants of integration.

## 2. How do I determine if a rational function can be rewritten as partial fractions?

A rational function can be rewritten as partial fractions if the degree of the numerator is less than the degree of the denominator. Additionally, if the denominator can be factored into linear and quadratic factors, it can be rewritten as partial fractions.

## 3. Can the integral of a rational function be solved with substitution?

Yes, the integral of a rational function can sometimes be solved with substitution. This typically involves substituting a variable for the denominator of the function and then using the power rule for integration.

## 4. Is there a shortcut method for integrating rational functions?

There is no shortcut method for integrating all rational functions. However, there are certain types of rational functions that have specific integration techniques, such as trigonometric substitutions or using the method of partial fractions.

## 5. Are there any special cases to consider when integrating rational functions?

Yes, there are a few special cases to consider when integrating rational functions. One is when the degree of the numerator is equal to the degree of the denominator, in which case long division can be used to rewrite the function. Another case is when the denominator has repeated linear factors, in which case a modified version of partial fractions may be needed.

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