一个双曲函数的积分

[Largedisplaystyle int^{infty}_{0}frac{ anhleft(\, x\, ight)} {xleft[\, 1 - 2coshleft(\, 2x\, ight)\, ight]^{2}}\,{ m d}x]

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(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*}]