[Largedisplaystyle int_0^inftyfrac{lnleft(1+x+sqrt{x^2+2\,x}
ight)\,lnleft(1+sqrt{x^2+2\,x+2}
ight)}{x^2+2x+1}mathrm dx
]
(Largemathbf{Solution:})
This integral can be solved by brutal force.
[I = int_{1}^{infty} frac{ln(x + sqrt{x^{2} - 1}) ln(1 + sqrt{x^{2} + 1})}{x^{2}} \, mathrm dx
]
By the knowledge in trigonometry, we obtain
[egin{align*}
I
&= int_{1}^{infty} frac{operatorname{arcosh} x cdot (operatorname{arsinh}(1/x) + ln x)}{x^{2}} \, mathrm dx \
&= int_{0}^{1} operatorname{arcosh} (1/x) cdot (operatorname{arsinh} x - ln x) \, mathrm dx
end{align*}]
Note that
[int (operatorname{arsinh} x - ln x) \, mathrm dx = xoperatorname{arsinh} x - x ln x + x + 1 - sqrt{x^{2} + 1}
]
Here, the constant of integration is chosen so that the integrand becomes (O(xln x)) as (x searrow 0). Since (operatorname{arcosh}(1/x) = O(ln x)) as (x searrow 0), we can perform integration by parts to obtain
[I = int_{0}^{1} left( operatorname{arsinh} x - ln x + 1 + frac{1 - sqrt{x^{2} + 1}}{x}
ight) \, frac{mathrm dx}{sqrt{1-x^{2}}}
]
Plug (x=sin heta). Then
[I = int_{0}^{frac{pi}{2}} left( operatorname{arsinh} (sin heta) - ln sin heta + 1 + frac{1 - sqrt{1 + sin^{2} heta}}{sin heta}
ight) \, mathrm d heta
]
We divide the integral into 4 parts and consider them separately.
Part 1. By the Taylor expansion of (operatorname{arsinh} x),
[egin{align*}
int_{0}^{frac{pi}{2}} operatorname{arsinh} (sin heta) \, mathrm d heta
&= sum_{n=0}^{infty} inom{-1/2}{n} frac{1}{2n+1} int_{0}^{frac{pi}{2}} sin^{2n+1} heta \, mathrm d heta \
&= frac{1}{2} sum_{n=0}^{infty} frac{1}{2n+1} frac{Gammaleft ( dfrac{1}{2}
ight )}{Gamma(n+1)Gammaleft ( dfrac{1}{2}
ight )} frac{Gamma(n+1)Gammaleft ( dfrac{1}{2}
ight )}{Gammaleft ( dfrac{3}{2}+n
ight )} \
&= sum_{n=0}^{infty} frac{(-1)^{n}}{(2n+1)^{2}}
\&= extbf{G}
end{align*}]
Part 2. We know so many ways to prove that
[-int_{0}^{frac{pi}{2}} ln sin heta \, mathrm d heta = frac{pi}{2}ln 2
]
Part 3. Do you need an explanation for this?
[int_{0}^{frac{pi}{2}} mathrm d heta = frac{pi}{2}
]
Part 4. Again, by the Taylor expansion,
[egin{align*}
int_{0}^{frac{pi}{2}} frac{1 - sqrt{1 + sin^{2} heta}}{sin heta} \, mathrm d heta
&= - sum_{n=1}^{infty} inom{1/2}{n} int_{0}^{frac{pi}{2}} sin^{2n-1} heta \, mathrm d heta \
&= - frac{1}{2} sum_{n=1}^{infty} frac{Gammaleft ( displaystylefrac{3}{2}
ight )}{Gamma(n+1)Gammaleft ( displaystylefrac{3}{2}-n
ight )} frac{Gamma(n)Gammaleft ( displaystylefrac{1}{2}
ight )}{Gammaleft ( displaystylefrac{1}{2}+n
ight )} \
&= sum_{n=0}^{infty} frac{(-1)^{n}}{(2n)(2n-1)}
\&= frac{ln 2}{2} - frac{pi}{4}
end{align*}]
Putting all these together, we obtain
[oxed{displaystyle int_0^inftyfrac{lnleft(1+x+sqrt{x^2+2\,x}
ight)\,lnleft(1+sqrt{x^2+2\,x+2}
ight)}{x^2+2x+1}mathrm dx=color{Blue} { extbf{G}+fracpi4+frac{pi+1}2ln2}}
]