Partial Fractions Is A Real Pain!

Relevant wiki: Integration Tricks

Here's the solution, but later on I'll discuss the process behind this.

Call our integral $I.$ Substitute $y= \dfrac{1}{x}$ (so $dy = -\dfrac{1}{x^2}dx$ ) and we get $I = \int_\infty^0 \frac{\frac{1}{x^2}}{\left(1+\frac{1}{x}\right) \left(1+\frac{1}{x^6}\right)}\cdot \left(-\frac{1}{x^2}\right)\,dx = \int_0^\infty \frac{\frac{1}{x^4}}{\left(1+\frac{1}{x}\right) \left(1+\frac{1}{x^6}\right)}\,dx = \int_0^\infty \frac{x^3}{(1+x)(1+x^6)}\,dx.$

Note that the bounds changed from $\infty$ to 0, but we then reversed them back and got rid of a negative sign. Adding this to our original integral, $I,$ gives $2I = \int_0^\infty \frac{y^2+y^3}{(1+y)(1+y^6)}\,dy = \int_0^\infty \frac{y^2}{1+y^6}\,dy.$ For this addition, it's probably worth showing that the integrals converge, but since they're equal and the sum ends up being finite, let's not worry about this.

Now, substitute $u = y^3$ (so $du = 3y^2dy$ ) giving $2I = \int_0^\infty \frac{1}{3}\cdot \frac{1}{1+u^2} \,du = \left.\frac{1}{3}\tan^{-1}(u)\right|_{0}^\infty = \frac{1}{3} \left(\frac{\pi}{2} - 0\right) = \frac{\pi}{6}.$ Thus, $I = \dfrac{1}{2}\cdot \dfrac{\pi}{6} = \dfrac{\pi}{12}.$

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If that looks like a bunch of magic (it should!), here's the problem-solving process behind this solution (for me):

I wanted to simplify things, and study related integrals. First, to simplify things for the purposes of exploring, I tried changing $y^2$ in the numerator to just a $1.$ When integrals go from 0 to $\infty,$ nice things often happen when you “reverse them” by using a $\frac{1}{x}$ substitution (to go from $\infty$ to 0); this is especially true with rational functions of polynomials. So, I noticed that this substitution gave the same integral except with $y^5$ in the numerator instead of $1.$ Interesting…

In retrospect, I could have just used the $\frac{1}{x}$ observation on the integral with $y^2$ in the numerator, but now I was curious: what happens if I have $y^k$ in the numerator? It turns out, via this substitution, that it equals the same integral with $y^{5-k}$ in the numerator!

So the $y^2$ integral equals the $y^3$ integral, but how does this help? I really hate having two terms in the denominator, and I’d love to get rid of one. Which one? Well…subtracting the integrals makes a numerator of $y^3 - y^2 = y^2 (y-1)$ which is pretty unhelpful, but $y^3 + y^2 = y^2(1+y)$ looks great -- one of the denominator terms will cancel with the numerator if I add the integrals!

The last substitution of $u=y^3$ is a more typical $u$ -substitution. I don’t want the numerator there, so I try to make it the derivative of something like $y^3.$ That works nicely, since the denominator is $1+(y^3)^2.$

Let y^3 = tanx Then 1+y^6 = (secx)^2 Also, 3y^2dy=(secx)^2dx. y--->0, tanx--->0, so x--->arctan0=0 y--->inf, tanx--->inf, x---->arctan(inf)=pi/2 Reduce and express resulting expression in terms of sines and cosines. Then use the instructions in the link below https://m.facebook.com/story.php?story_fbid=489280981275670&id=309749225895514 I hope posting the link here does not violate rules. Otherwise, delete the comment