[Largedisplaystyle int_{0}^{1}frac{lnleft ( x+sqrt{x^{2}+1}
ight )}{x}mathrm{d}x
]
(Largemathbf{Solution:})
注意到：
[int_{0}^{1}frac{lnleft ( x+sqrt{x^{2}+1}
ight )}{x}mathrm{d}x=int_{0}^{1}frac{mathrm{arcsinh} x}{x}mathrm{d}x
]
分部即得：
[int_{0}^{1}frac{mathrm{arcsinh} x}{x}mathrm{d}x=mathrm{arcsinh} xln xBigg|_{0}^{1}-int_{0}^{1}frac{ln x}{sqrt{1+x^{2}}}mathrm{d}x
]
令
[mathcal{I}left ( alpha
ight )=int_{0}^{1}frac{x^{alpha }}{sqrt{1+x^{2}}}mathrm{d}x
]
我们有
[egin{align*}
mathcal{I}left ( alpha
ight )&=int_{0}^{1}frac{x^{alpha }}{sqrt{1+x^{2}}}mathrm{d}x=int_{0}^{1}x^{alpha } left ( 1+x^{2}
ight )^{-frac{1}{2}}mathrm{d}x\
&=frac{1}{2}int_{0}^{1}x^{frac{alpha }{2}-frac{1}{2}}left ( 1+x
ight )^{-frac{1}{2}}mathrm{d}x\
&=frac{1}{1+alpha }\, _{2}F_{1}left ( frac{1}{2},frac{1+alpha }{2};frac{3+alpha }{2};-1
ight )
end{align*}]
所以
[largeoxed{displaystyle egin{align*}
int_{0}^{1}frac{lnleft ( x+sqrt{x^{2}+1}
ight )}{x}mathrm{d}x&=-int_{0}^{1}frac{ln x}{sqrt{1+x^{2}}}mathrm{d}x=-mathcal{I}'left ( 0
ight )\
&=color{blue}{mathrm{arcsinh}left ( 1
ight )-frac{1}{2}\, _{2}F_{1}^{left ( 0,1,0,0
ight )}left ( frac{1}{2},frac{1}{2};frac{3}{2};-1
ight )}\[8pt]
&~~~color{blue}{-frac{1}{2}\, _{2}F_{1}^{left ( 0,0,1,0
ight )}left ( frac{1}{2},frac{1}{2};frac{3}{2};-1
ight )}\[8pt]
&approx 0.95520180648118
end{align*}}]