<s>[INST] How to find the exact value of the integral $ \int_{0}^{\infty} \frac{d x}{\left(x^{3}+\frac{1}{x^{3}}\right)^{2}}$? $\textrm{I first reduce the power two to one by Integration by Parts.}$
$\begin{aligned}\displaystyle \int_{0}^{\infty} \frac{1}{\left(x^{3}+\frac{1}{x^{3}}\right)^{2}} d x &=\int_{0}^{\infty} \frac{x^{6}}{\left(x^{6}+1\right)^{2}} d x\\&=\displaystyle -\frac{1}{6} \int_{0}^{\infty} x d\left(\frac{1}{x^{6}+1}\right)\\&
=\displaystyle -\left[\frac{x}{6\left(x^{6}+1\right)}\right]_{0}^{\infty}+\frac{1}{6} \int_{0}^{\infty} \frac{1}{x^{6}+1} d x \quad \textrm{ (Via Integration by Parts})\\&=\displaystyle \frac{1}{6} \int_{0}^{\infty} \frac{1}{x^{6}+1} d x\end{aligned}$
$\textrm{Then I am planning to evaluate }\displaystyle I= \int_{0}^{\infty} \frac{1}{x^{6}+1}\text{ by resolving }\frac{1}{x^{6}+1} \text{ into partial fractions.}$
But after noticing that $$I=\int_{0}^{\infty} \frac{d x}{x^{6}+1}\stackrel{x\mapsto\frac{1}{x}}{=} \int_{0}^{\infty} \frac{x^{4}}{x^{6}+1} d x,$$
I changed my mind and started with $3I$ instead of $I$ as below:
$$
\begin{aligned}
3 I &=\int_{0}^{\infty} \frac{x^{4}+2}{\left(x^{2}+1\right)\left(x^{4}-x^{2}+1\right)} d x \\
&=\int_{0}^{\infty}\left(\frac{1}{x^{2}+1}+\frac{1}{x^{4}-x^{2}+1}\right) d x \\
&=\left[\tan ^{-1} x\right]_{0}^{\infty}+\int_{0}^{\infty} \frac{\frac{1}{x^{2}}}{x^{2}+\frac{1}{x^{2}}-1} d x \\
&=\frac{\pi}{2}+\frac{1}{2} \int_{0}^{\infty} \frac{\left(1+\frac{1}{x^{2}}\right)-\left(1-\frac{1}{x^{2}}\right)}{x^{2}+\frac{1}{x^{2}}-1} d x\\ 
&=\frac{\pi}{2}+\frac{1}{2}\left[\int_{0}^{\infty} \frac{d\left(x-\frac{1}{x}\right)}{\left(x-\frac{1}{x}\right)^{2}+1}-\int \frac{d\left(x+\frac{1}{x}\right)}{\left(x+\frac{1}{x}\right)^{2}-3}\right] \\
&=\frac{\pi}{2}+\frac{1}{2}\left[\tan ^{-1}\left(x-\frac{1}{x}\right)\right]_{0}^{\infty}-0 \\
&=\frac{\pi}{2}+\frac{1}{2}\left[\frac{\pi}{2}-\left(-\frac{\pi}{2}\right)\right] \\
&=\pi \\ \therefore I &=\frac{\pi}{3}
\end{aligned}
$$
Now I can conclude that
$$\boxed{\displaystyle \quad \int_{0}^{\infty} \frac{1}{\left(x^{3}+\frac{1}{x^{3}}\right)^{2}} d x=\frac{\pi}{18} }.$$
:|D Wish you enjoy the solution! Opinions and alternative  methods are welcome.
 [/INST] It is a nice solution for sure.
As you showed the problem is to compute
$$I=\int \frac {dx}{x^6+1}$$
My working
Writing
$$x^6+1=(x^2+1)(x^4-x^2+1)=(x^2+1)(x^2-a)(x^2-b)$$  Using partial fraction decomposition
$$\frac 1 {(x^2+1)(x^2-a)(x^2-b)}=$$ $$\frac{1}{(a+1) (a-b) \left(x^2-a\right)}-\frac{1}{(b+1) (a-b)
   \left(x^2-b\right)}+\frac{1}{(a+1) (b+1) \left(x^2+1\right)}$$ Integrating
$$I=\frac{\tan ^{-1}(x)}{(a+1) (b+1)}-\frac 1{a-b}\Bigg[\frac{\tanh ^{-1}\left(\frac{x}{\sqrt{a}}\right)}{(a+1)\sqrt{a} }-\frac{\tanh
   ^{-1}\left(\frac{x}{\sqrt{b}}\right)}{(b+1)\sqrt{b} } \Bigg]$$ Using the bounds, we then have
$$J=\int_0^\infty \frac {dx}{x^6+1}=\frac{\pi }{2 (a+1) (b+1)}+\frac \pi{2(a-b)}\Bigg[\frac{\sqrt{-\frac{1}{a}}}{a+1}-\frac{\sqrt{-\frac{1}{b}}}{b+1}\Bigg]$$
$$a=\frac{1+i\sqrt{3}}{2}=   \sqrt{-\frac{1}{a}}\qquad \text{and} \qquad
b=\frac{1-i\sqrt{3}}{2}=   \sqrt{-\frac{1}{b}} $$ So, finally
$$J=\frac{\pi }{1+(a+b)+a b}=\frac \pi 3$$
</s>