Neither square nor cube root. It's 10th!

The integral $\int \limits^{\infty }_{0}\frac{x^{\frac{1}{10} }}{x^{2}+1} dx= \frac{1}{2} \int \limits^{\infty }_{0}\frac{u^{\frac{-9}{20} }}{u+1} du \ \ \ \ \ \ \ (1)$ $= \ \ \ \ \frac{1}{2} \int \limits^{\infty }_{0}\frac{u^{\frac{-9}{20} }}{u+1} du =\frac{\beta \left( 1-\frac{9}{20} , \frac{9}{20} \right) }{2} \ \ \ \ \ \ (2)$ $=\frac{\Gamma \left( 1-\frac{9}{20} \right) \Gamma \left( \frac{9}{20} \right) }{2} = \frac{\pi }{2\sin \left( \frac{9\pi }{20} \right)} \ \ \ \ \ \ \ \ (3)$ $\frac{\pi }{2\sin \left( \frac{9\pi }{20} \right) } =\frac{\pi }{2} \sec \left( \frac{\pi }{20} \right)$ So $A+B+C+D= \boxed{24}$

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$(1)$ use the substitute $x^2=u$

$(2)$ use the definition of beta function $\beta \left( x,y\right) =\int \limits^{\infty }_{0}\frac{t^{x-1}}{\left( 1+t\right) ^{x+y}} dt$

$(3)$ use Euler reflection formula $\Gamma \left( x\right) \Gamma \left( 1-x\right) =\frac{\pi }{\sin \left( \pi x\right) }$