Integral using Beta and Gamma functions

Question:
Find the value of
\[ \int^{\infty}_{-\infty}\frac{dx}{x^6+1} \]

Answer:
We can reduce the above integral into
\[ \int^{\infty}_{-\infty}\frac{dx}{x^6+1} = 2\int^{\infty}_{0}\frac{dx}{x^6+1} \]
Replace $x$ with $t^{\frac{1}{6}}$. Then, we have $dx=\frac{1}{6}t^{-\frac{5}{6}}dt$. Thus,
\[ \int^{\infty}_{0}\frac{dx}{x^6+1} = \frac{1}{6}\int^{\infty}_{0}\frac{t^{-\frac{5}{6}}}{t+1}dt \]
Remember that the Beta function is defined as follows:
\[ B(p,q) = \int^{\infty}_{0}\frac{x^{p-1}}{(1+x)^{p+q}}dx \\
B(p,q)=B(q,p)=\frac{\Gamma(p)\Gamma(q)}{\Gamma(p+q)} \]
Therefore, the above expression becomes
\[ \frac{1}{6}\int^{\infty}_{0}\frac{t^{-\frac{5}{6}}}{t+1}dt = \frac{1}{6}B(\frac{1}{6},\frac{5}{6}) \\
   = \frac{1}{6}\frac{\Gamma(\frac{1}{6})\Gamma(\frac{5}{6})}{\Gamma(\frac{1}{6}+\frac{5}{6})} \\
   =  \frac{1}{6}\Gamma(\frac{1}{6})\Gamma(\frac{5}{6})  \]
since $\Gamma(1)=1$.
Recall the following formula:
\[ \Gamma(s)\Gamma(1-s) = \frac{\pi}{\sin\pi s}  \]
The parameter satisfies 0<s<1.
Therefore,
\[ \frac{1}{6}\Gamma(\frac{1}{6})\Gamma(\frac{5}{6}) = \frac{1}{6}\frac{\pi}{\sin(\pi/6)} = \frac{\pi}{3} \]
Hence,
\[ \int^{\infty}_{-\infty}\frac{dx}{x^6+1} = \frac{2\pi}{3}\]

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