[Largedisplaystyle int^{infty}_{0}frac{ anhleft(\, x\,
ight)}
{xleft[\, 1 - 2coshleft(\, 2x\,
ight)\,
ight]^{2}}\,{
m d}x]
(Largemathbf{Solution:})
A possible way I see of doing this is to apply the substitution (xmapsto-ln{x}), which yields
[-int^{1}_{0}frac{x^{3}left(\, 1 - x^{2}\,
ight)}
{left(\, 1 + x^{2}\,
ight)left(\, 1 - x^{2} + x^{4}\,
ight)^{2}}\,
frac{{
m d}x}{lnleft(\, x\,
ight)}]
So
[egin{align*}
mathcal{I}
&=-int_{0}^{1}frac{x^3(1-x^2)}{(1+x^2)(1-x^2+x^4)^2}frac{mathrm{d}x}{ln{x}}\&=-int_{0}^{1}frac{x^3(1-x^2)}{(1+x^6)(1-x^2+x^4)}frac{mathrm{d}x}{ln{x}}\
&=-int_{0}^{1}frac{x^3(1-x^4)}{(1+x^6)^2}frac{mathrm{d}x}{ln{x}}\&=-int_{0}^{1}frac{z(1-z^{4/3})}{(1+z^2)^2}frac{mathrm{d}z}{3z^{2/3}ln{left(z^{1/3}
ight)}}\&=int_{0}^{1}\,frac{1}{(1+z^2)^2}frac{z^{5/3}-z^{1/3}}{ln{z}}mathrm{d}z\
&=int_{0}^{1}\,frac{mathrm{d}z}{(1+z^2)^2}int_{1/3}^{5/3}\,z^{mu}mathrm{d}mu\&=int_{1/3}^{5/3}mathrm{d}muint_{0}^{1}\,frac{z^{mu}}{(1+z^2)^2}mathrm{d}z\&=int_{1/3}^{5/3}left[-frac14+frac{mu-1}{4}eta{left(frac{mu-1}{2}
ight)}
ight]mathrm{d}mu\
&=-frac13+int_{1/3}^{5/3}left[frac{mu-1}{4}eta{left(frac{mu-1}{2}
ight)}
ight]mathrm{d}mu\&=-frac13+int_{-1/3}^{1/3}\,teta{left(t
ight)}mathrm{d}t\
&=-frac13+int_{-1/3}^{1/3}\,frac{t}{2}left[psi{left(frac{t+1}{2}
ight)}-psi{left(frac{t}{2}
ight)}
ight]mathrm{d}t\&=-frac13+int_{-1/3}^{1/3}\,frac{t}{2}psi{left(frac{t+1}{2}
ight)}mathrm{d}t-int_{-1/3}^{1/3}\,frac{t}{2}psi{left(frac{t}{2}
ight)}mathrm{d}t\
&=-frac13+int_{1/3}^{2/3}\,(2u-1)psi{left(u
ight)}mathrm{d}u-2int_{-1/6}^{1/6}\,upsi{left(u
ight)}mathrm{d}u\
&=-frac13-int_{1/3}^{2/3}\,psi{left(u
ight)}mathrm{d}u+2int_{1/3}^{2/3}\,upsi{left(u
ight)}mathrm{d}u-2int_{-1/6}^{1/6}\,upsi{left(u
ight)}mathrm{d}u\
&=-frac13+ln{left(frac{Gamma{left(dfrac13
ight)}}{Gamma{left(dfrac23
ight)}}
ight)}+2int_{1/3}^{2/3}\,upsi{left(u
ight)}mathrm{d}u-2int_{-1/6}^{1/6}\,upsi{left(u
ight)}mathrm{d}u\
&=-frac13+ln{left(frac{Gamma{left(dfrac13
ight)}}{Gamma{left(dfrac23
ight)}}
ight)}+2int_{1/3}^{2/3}\,upsi{left(u
ight)}mathrm{d}u-2int_{5/6}^{7/6}\,(1-v)psi{left(1-v
ight)}mathrm{d}v\
&=-frac13+ln{left(frac{Gamma{left(dfrac13
ight)}}{Gamma{left(dfrac23
ight)}}
ight)}+2left[uln{Gammaleft(u
ight)}-psi^{(-2)}{left(u
ight)}
ight]_{1/3}^{2/3}\
&~~~~~ +2left[(1-v)ln{Gammaleft(1-v
ight)}-psi^{(-2)}{left(1-v
ight)}
ight]_{5/6}^{7/6}\
&=ln{left(frac{Gamma{left(dfrac13
ight)}}{Gamma{left(dfrac23
ight)}}
ight)}-frac{5pi}{9sqrt{3}}-frac{ln{left(2pi
ight)}}{3}+frac23ln{left(frac{Gamma{left(dfrac23
ight)}^2}{Gamma{left(dfrac13
ight)}}
ight)}+frac{5psi^{(1)}{left(dfrac13
ight)}}{6sqrt{3}\,pi}\
&=Largeoxed{displaystylecolor{blue}{-frac{5pi}{9sqrt{3}}-frac{ln{3}}{6}+frac{5psi^{(1)}{left(dfrac13
ight)}}{6sqrt{3}\,pi}}}
end{align*}]