Integral $\int_0^\infty\frac{\ln\left(1+x+\sqrt{x^2+2\,x}\right)\,\ln\left(1+\sqrt{x^2+2\,x+2}\right)}{x^2+2x+1}dx$

Could you suggest any ideas how to evaluate this integral? Is there a closed-form result? $$\int_0^\infty\frac{\ln\left(1+x+\sqrt{x^2+2\,x}\right)\,\ln\left(1+\sqrt{x^2+2\,x+2}\right)}{x^2+2x+1}dx$$

This integral can be solved by brutal force. I will use Vladimir's suggestion as a starting point:

$$I = \int_{1}^{\infty} \frac{\log(x + \sqrt{x^{2} - 1}) \log(1 + \sqrt{x^{2} + 1})}{x^{2}} \, dx. $$

By the knowledge in trigonometry, we obtain

\begin{align*} I &= \int_{1}^{\infty} \frac{\operatorname{arcosh} x \cdot (\operatorname{arsinh}(1/x) + \log x)}{x^{2}} \, dx \\ &= \int_{0}^{1} \operatorname{arcosh} (1/x) \cdot (\operatorname{arsinh} x - \log x) \, dx. \end{align*}

Note that

$$ \int (\operatorname{arsinh} x - \log x) \, dx = x\operatorname{arsinh} x - x \log x + x + 1 - \sqrt{x^{2} + 1}. $$

Here, the constant of integration is chosen so that the integrand becomes $O(x\log x)$ as $x \searrow 0$. Since $\operatorname{arcosh}(1/x) = O(\log x)$ as $x \searrow 0$, we can perform integration by parts to obtain

$$ I = \int_{0}^{1} \left( \operatorname{arsinh} x - \log x + 1 + \frac{1 - \sqrt{x^{2} + 1}}{x} \right) \, \frac{dx}{\sqrt{1-x^{2}}}. $$

Plug $x = \sin\theta$. Then

$$ I = \int_{0}^{\frac{\pi}{2}} \left( \operatorname{arsinh} (\sin\theta) - \log \sin\theta + 1 + \frac{1 - \sqrt{1 + \sin^{2}\theta}}{\sin\theta} \right) \, d\theta. $$

We divide the integral into 4 parts and consider them separately.

Part 1. By the Taylor expansion of $\operatorname{arsinh} x$,

\begin{align*} \int_{0}^{\frac{\pi}{2}} \operatorname{arsinh} (\sin\theta) \, d\theta &= \sum_{n=0}^{\infty} \binom{-1/2}{n} \frac{1}{2n+1} \int_{0}^{\frac{\pi}{2}} \sin^{2n+1}\theta \, d\theta \\ &= \frac{1}{2} \sum_{n=0}^{\infty} \frac{1}{2n+1} \frac{\Gamma(\frac{1}{2})}{\Gamma(n+1)\Gamma(\frac{1}{2}-n)} \frac{\Gamma(n+1)\Gamma(\frac{1}{2})}{\Gamma(\frac{3}{2}+n)} \\ &= \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n+1)^{2}} = G. \end{align*}

Part 2. We know so many ways to prove that

$$ -\int_{0}^{\frac{\pi}{2}} \log \sin\theta \, d\theta = \frac{\pi}{2}\log 2. $$

Part 3. Do you need an explanation for this?

$$ \int_{0}^{\frac{\pi}{2}} d\theta = \frac{\pi}{2}. $$

Part 4. Again, by the Taylor expansion,

\begin{align*} \int_{0}^{\frac{\pi}{2}} \frac{1 - \sqrt{1 + \sin^{2}\theta}}{\sin\theta} \, d\theta &= - \sum_{n=1}^{\infty} \binom{1/2}{n} \int_{0}^{\frac{\pi}{2}} \sin^{2n-1}\theta \, d\theta \\ &= - \frac{1}{2} \sum_{n=1}^{\infty} \frac{\Gamma(\frac{3}{2})}{\Gamma(n+1)\Gamma(\frac{3}{2}-n)} \frac{\Gamma(n)\Gamma(\frac{1}{2})}{\Gamma(\frac{1}{2}+n)} \\ &= \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n)(2n-1)} = \frac{\log 2}{2} - \frac{\pi}{4}. \end{align*}

Putting all these together, we obtain the answer as Cleo claimed.

This is a very interesting integral! I'll give you a partial answer:
Yes, there is a closed form expression for it: $$G+\frac\pi4+\frac{\pi+1}2\ln2,$$ where $G$ is the Catalan constant: $$G=-\int_0^1\frac{\ln x}{x^2+1}dx.$$