How to calculate the integral $\int_{a}^{b} \frac{\ln x}{\sqrt{x^{2}+1}} \, \mathrm dx $ 
For some $a$, $b$ in the domain of the integration of $f(x)=
 \frac{\ln x}{\sqrt{x^{2}+1}}$, calculate $I$.
$$I=\int_{a}^{b}\frac{\ln x}{\sqrt{x^2+1}} \, \mathrm{d}x$$

But how do you find a primitive function for $\frac{\ln x}{\sqrt{x^2+1}}$? It doesn't appear to be something that can be expressed in terms of a finite number of elementary functions.
 A: Under the assumption that $x>0$,
$$ \begin{align} \int \frac{\ln x}{\sqrt{1+x^{2}}} \, dx &= \int \ln (\sinh u) \, du \\ &=\int \ln\left( \frac{e^{u}-e^{-u}}{2} \right) \, du \\ &=\int \ln \left( \frac{e^{u} (1-e^{-2u})}{2} \right) \ du \\ &= \int \ln(e^{u})\, du +\int \ln(1-e^{-2u}) \, du- \ln 2 \int  \, du \\ &=\int u \, du -\frac{1}{2} \int \frac{\ln(1-w)}{w} \, dw -u \ln 2  \\ &= \frac{u^{2}}{2} + \frac{\operatorname{Li}_{2}(w)}{2} - u \ln 2 + C \tag{1}\\ &= \frac{u^{2}}{2} + \frac{\operatorname{Li}_{2} (e^{-2u})}{2}- u \ln 2 +C \\& = \frac{\operatorname{arsinh}^{2}(x)}{2} + \frac{\operatorname{Li}_{2} (e^{-2 \operatorname{arsinh} x})}{2}- \operatorname{arsinh}(x) \ln 2 + C \end{align} $$
$(1)$ https://en.wikipedia.org/wiki/Polylogarithm#Dilogarithm
The antiderivative provided by Wolfram Alpha is valid over a larger domain, but apparently it simplifies to this result if $x$ is assumed to be positive.
A: $\newcommand{\+}{^{\dagger}}
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We'll assume $\ds{0 < a < b}$:

$$
I=\int_{a}^{b}{\ln\pars{x} \over \root{x^{2} + 1}}\,\dd x
=\int_{\theta_{a}}^{\theta_{b}}{\ln\pars{\sinh\pars{\theta}}}\,\dd\theta
\quad\mbox{where}\quad\theta_{\mu} = {\rm arcsinh}\pars{\mu}\,,\quad \mu = a, b.
$$

\begin{align}
\int\ln\pars{\sinh\pars{\theta}}
=
\half\left\{%
\text{Li}_2\left({\rm e}^{-2\theta}\right)
-\theta\left[\theta + 2 \ln\left(1 - {\rm e}^{-2\theta}\right)
-2\ln\left(\sinh\left(\theta\right)\right)\right]\right\}
\end{align}
A: You can find the integrals fairly easily for most derivatives and integrals using WolframAlpha.  
The equation that you would enter into Wolfram would look like this:
int(log(x)/sqrt(x^2+1))

Because of the computational complexity, Wolfram wants you to pay for the "pro" service, but you could just evaluate the output.  Here is a link to the Wolfram Output.
