Evaluation of $\int_{0}^{\infty}\frac{1}{\sqrt{x^4+x^3+x^2+x+1}}dx$ Evaluation of Integral $\displaystyle \int_{0}^{\infty}\frac{1}{\sqrt{x^4+x^3+x^2+x+1}}dx$
$\bf{My\; Try::}$ First we will convert $x^4+x^3+x^2+x+1$ into closed form, which is $\displaystyle \left(\frac{x^5-1}{x-1}\right)$
So Integral is $\displaystyle \int_{0}^{\infty}\frac{\sqrt{x-1}}{\sqrt{x^5-1}}dx$
now i did not understand how can i solve it
Help me
Thanks

$$
I=\int^{\infty}_{0}\frac{dx}{\sqrt{x^4+x^3+x^2+x+1}}
$$
Making the change of variable $x\rightarrow y/(1+y)$ we get
$$
I=\int^{1}_{0}\frac{dy}{\sqrt{y^4-2y^3+4y^2-3y+1}}=
$$
$$
=\int^{1}_{0}\frac{dy}{\sqrt{1/16(2y-1)^4+5/8(2y-1)^2+5/16}}.
$$
Also setting $(2y-1)\rightarrow u$, then
$$
I=\frac{1}{2}\int_{-1}^{1}\frac{du}{\sqrt{1/16 (u^2+5)^2-5/4}}=\int_{0}^{1}\frac{du}{\sqrt{1/16 (u^2+5)^2-5/4}}
$$
Then setting $(u^2+5)/4\rightarrow w$, we get
$$
I=\int_{5/4}^{6/4}\frac{dw}{\sqrt{w^2-5/4}}\frac{1}{\sqrt{w-5/4}}=\int_{0}^{1/4}\frac{w^{-1/2}dw}{\sqrt{(w+5/4-\sqrt{5/4})(w+5/4+\sqrt{5/4})}}=
$$
$$
=-2i\frac{5-\sqrt{20}}{\sqrt{5}\sqrt{5-\sqrt{20}}}\int^{-1/(5-\sqrt{20})}_{0}\frac{t^{-1/2}}{\sqrt{(1-t)(1-mt)}}dt,
$$
where $m=\frac{5-\sqrt{20}}{5+\sqrt{20}}$. From the deffinition of the elliptic integrals (in Mathematica's notation):
$$
F(\sin\phi,m)=\int_{0}^{\phi}\frac{1}{\sqrt{(1-t^2)(1-mt^2)}}dt=\frac{1}{2}\int_{0}^{ \phi^2}\frac{1}{\sqrt{t(1-t)(1-mt)}}dt\Leftrightarrow
$$
$$
2F\left(\sin\sqrt{\phi},m\right)=\int^{ \phi}_{0}\frac{1}{\sqrt{t(1-t)(1-mt)}}dt.
$$
Hence from (1) and (2) we have
$$
I=-4i\frac{5-\sqrt{20}}{\sqrt{5}\sqrt{5-\sqrt{20}}}\cdot F\left(\arcsin\left(\frac{i}{\sqrt{5-\sqrt{20}}}\right),\frac{5-\sqrt{20}}{5+\sqrt{20}}\right).
$$
