TITLE: How to calculate the integral $\int_{a}^{b} \frac{\ln x}{\sqrt{x^{2}+1}} \, \mathrm dx $
QUESTION [3 upvotes]: 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.

REPLY [6 votes]: 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.
{"set_name": "stack_exchange", "score": 3, "question_id": 738874}