Twelve? Ain't Nobody got time for that! Part 2

Use the substitution $x^{12}=t \Rightarrow dx=\frac{1}{12}t^{-11/12}\,dt$ to obtain:

$\int_0^{\infty} \frac{dx}{x^{12}+1}=\frac{1}{12}\int_0^{\infty} \frac{t^{-11/12}}{t+1}\,dt$

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Now I prove the following result:

$\int_0^\infty \dfrac{x^{-a}}{x+1}\mathrm{d}x=\frac{\pi}{\sin(\pi a)}$

Write the integral as:

$\displaystyle \int_0^{\infty} \int_0^{\infty} x^{-a} e^{-(x+1)t}\,dt\,dx=\int_0^{\infty} \left(\int_0^{\infty} x^{-a}e^{-xt}\,dx\right) e^{-t}\,dt$

First I evaluate:

$\displaystyle \int_0^{\infty} x^{-a}e^{-xt}\,dx$

Use the substitution $xt=y$ to obtain:

$\frac{1}{t^{-a+1}} \int_0^{\infty} y^{-a} e^{-y}\,dy=\frac{1}{t^{1-a}}\Gamma(1-a)$

Hence,

$\int_0^{\infty} \left(\int_0^{\infty} x^{-a}e^{-xt}\,dx\right) e^{-t}\,dt=\Gamma(1-a)\int_0^{\infty} t^{a-1}e^{-t}\,dt=\Gamma(1-a)\Gamma(a)$

....which is by Euler's reflection formula:

$\Gamma(1-a)\Gamma(a)=\frac{\pi}{\sin(\pi a)}$

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Hence,

$\frac{1}{12}\int_0^{\infty} \frac{t^{-11/12}}{t+1}\,dt= \frac{\pi}{12\sin(\pi/12)}$

Since $\sin(\pi/12)=\frac{\sqrt{3}-1}{2\sqrt{2}}$ , the final answer is:

$\boxed{\dfrac{\pi (\sqrt{6}+\sqrt{2})}{12}}$

Another way & also to generalize the method, we can use substitution $\displaystyle y=\frac{1}{1+x^n}$ and the integral becomes Beta function $\frac{1}{n}\int_0^1 y^{\large 1-\frac{1}{n}-1}\ (1-y)^{\large \frac{1}{n}-1}\,dy=\frac{\Gamma\left(1-\frac{1}{n}\right)\Gamma\left(\frac{1}{n}\right)}{n}=\frac{\pi}{n\sin\frac{\pi}{n}}$ Try this one (click the problem):

$\int_0^{\Large\frac{\pi}{4}}\left[\frac{(1-x^2)\ln(1+x^2)+(1+x^2)-(1-x^2)\ln(1-x^2)}{(1-x^4)(1+x^2)}\right] x\, \exp\left[\frac{x^2-1}{x^2+1}\right]\, dx$ .

Alternative approach:

Substitute x = $(\tan y)^{1/12}$

then the integral will transform into trignometric form of Beta function.

Relevant wiki: Gamma Function

$\begin{aligned} I & = \int_0^\infty \frac 1{1+x^{12}}dx & \small \color{#3D99F6} \text{Let }u = x^{12} \implies du = 12 x^{11} dx \\ & = \frac 1{12} \int_0^\infty \frac {u^{-\frac {11}{12}}}{1+u} du & \small \color{#3D99F6} \text{Beta function }B(m,n) = \int_0^\infty \frac {u^{m-1}}{(1+u)^{m+n}} du \\ & = \frac 1{12} B \left(\frac 1{12}, \frac {11}{12} \right) & \small \color{#3D99F6} B(m,n) = \frac {\Gamma (m)\Gamma(n)}{\Gamma(m+n)} \text{, where }\Gamma (\cdot) \text{ is gamma function.} \\ & = \frac {\color{#3D99F6}\Gamma\left(\frac 1{12}\right)\Gamma\left(\frac {11}{12}\right)}{12 \color{#D61F06}\Gamma(1)} & \small \color{#3D99F6} \Gamma (x) \Gamma(1-x) = \frac \pi{\sin (\pi x)} \\ & = \frac {\color{#3D99F6}\frac \pi{\sin \frac \pi{12}}}{12 \cdot \color{#D61F06}0!} & \small \color{#D61F06} \Gamma (n) = (n-1)! \\ & = \frac {\pi(\sqrt 2+\sqrt 6)}{12} \end{aligned}$

Therefore, $a+b+c = 2+6+12 = \boxed{20}$ .

(x¹²+1)=((x⁴)³+1³)=(x⁴+1)(x^8-x⁴+1)=
(x⁴+1)(x⁴+√3x²+1)(x⁴-√3x²+1)
Now is easy 1/(x¹²+1)=
(ax³+bx²+cx+d)/(x⁴+1)+
(ex³+fx²+gx+h)/(x⁴+√3x²+1)+
(ix³+jx²+kx+l)/(x⁴-√3x²+1)
Now just solve that system and we get
a=b=c=e=g=i=k=0 , h=l=1/3,
f=-j=1/(2√3)
Now is easier
X⁴+1=(x²+x√2+1)(x²-x√2+1)
x⁴-√3x²+1=(x²+1)²-(bx)²=(x²-Bx+1)(x²+bx+1), where B=(1+√3)/√2
x⁴+√3x²+1=(x²-Ax+1)(x²+Ax+1) where A=(√3-1)/√2
Now is easy...