It's not like the others

$\displaystyle \int_{0}^{\infty}{\frac{1-{x}^{22}}{1-{x}^{24}} dx}$

$\displaystyle \int_{0}^{\infty}{\frac{1}{1-{x}^{24}} dx} - \int_{0}^{\infty}{\frac{{x}^{22}}{1-{x}^{24}} dx}$

$\displaystyle {I}_{1} = \int_{0}^{\infty}{\frac{1}{1-{x}^{24}} dx}, {I}_{2} = \int_{0}^{\infty}{\frac{{x}^{22}}{1-{x}^{24}} dx}$

$\displaystyle {I}_{1} : \quad {x}^{24}\rightarrow x$

$\displaystyle \frac{1}{24}\int_{0}^{\infty}{\frac{{x}^{-23/24}}{1-x} dx}$

$\displaystyle \frac{1}{24} M\left(\frac{1}{1-x}\right)\left(\frac{1}{24}\right)$

$M(f)(s)$ - Mellin transform of $f$ at $s$

Using Ramanujan Master Theorem,

$\displaystyle \frac{1}{24} \Gamma\left(\frac{1}{24}\right)\Gamma\left(1 - \frac{1}{24}\right){(-1)}^{1/24}$

Using Euler's Reflection Formula,

$\displaystyle \frac{1}{24} \frac{\pi}{sin\left(\frac{\pi}{24}\right)} {e}^{\pi\iota/24}$

$\displaystyle {I}_{2} : \quad {x}^{24}\rightarrow x$

$\displaystyle \frac{1}{24}\int_{0}^{\infty}{\frac{{x}^{-1/24}}{1-x} dx}$

$\displaystyle \frac{1}{24} M\left(\frac{1}{1-x}\right)\left(\frac{23}{24}\right)$

Proceeding as before,

$\displaystyle \frac{1}{24} \frac{\pi}{sin\left(\frac{23\pi}{24}\right)} {e}^{23\pi\iota/24}$

Putting them in one,

$\displaystyle \frac{\pi}{24 sin\left(\frac{\pi}{24}\right)}\left({e}^{\pi\iota/24} + {e}^{-\pi\iota/24}\right)$

$\displaystyle \frac{\pi}{12} cot\left(\frac{\pi}{24}\right)$

$\displaystyle \frac{\pi}{12} \left(2 + \sqrt{6} + \sqrt{3} + \sqrt{2}\right)$

$\displaystyle \frac{\pi}{\sqrt{12}}\left(\frac{1}{\sqrt{3}} + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{4}} + \frac{1}{\sqrt{6}}\right)$

I suspected this as a wrong direction however I know that I can solve via numerical method.