This Integral Will Surprise You

An absolutely filthy integral.

[Marius_Biggest_Fan]

Ok now derivate the answer. Lets see it turn back into 1/(x^6 +1)

[bn]

love the quick format also love "lawn" 😅

[maburwanemokoena]

To think a derivative and an integral are related but one is extremely hard compared to the other

[MichaelRothwell]

This is a nice exercise in integrating a rational function.

For me, the surprise was the way you factorised the quartic x⁴-x²+1:

=(x²+1+√3x)(x²+1-√3x)
=(x²+√3x+1)(x²-√3x+1)

How did you come up with this "trick"?

Here is my method (without using your trick) which is rather longer.

We note that x⁶+1=(x²+1)(x⁴-x²+1).
(This comes from the standard factorisation x³+1=(x+1)(x²-x+1), replacing x by x²)

On the other hand,
x⁶+1=0 ⇔x⁶=-1
which, as -1=cis(π), has solutions x=cis(±π/6), cis(±3π/6), cis(±5π/6).

cis(±3π/6)=cis(±π/2)=±i leading to factors x-i, x+i with product (x-i)(x+i)=x²+1.

So x=cis(±π/6), cis(±5π/6) are the solutions of x⁴-x²+1=0.

The roots cis(±π/6) have sum 2cos(π/6)=√3 and product 1, so satisfy the quadratic equation x²-√3x+1=0.

The roots cis(±5π/6) have sum 2cos(5π/6)=-2cos(π/6)=-√3 and product 1, so satisfy the quadratic equation x²+√3x+1=0.

Hence x⁴-x²+1=(x²+√3x+1)(x²-√3x+1).

Now, once I know this, I can write
(x²+√3x+1)(x²-√3x+1)

=x⁴+2x²+1-3x²=x⁴-x²+1
Now I can run the argument backwards to get what you wrote!

So my question is: did you also follow this route to arrive at your "trick", or is there some easy way to reach it?

[SamuRai-dedm]

had to retake a math course and ill probably just use this as my study guide. thanks vince.

[abdousekhi]

can we use partial fraction with complex root?