<s>[INST] Integral $\int_0^1 \frac{\ln(x+\sqrt{1-x^2})}{x}dx$ By integration by parts and the substitution $x = \sin t$ we can easily calculate the integral $\int_{0}^{1} \ln (x+ \sqrt{1-x^2})dx$ which equals to $\sqrt{2} \ln (\sqrt{2} +1) -1.$
I’ve tried to use the same substitution $x = \sin t$ to calculate the integral $ \int_{0}^{1} \frac {\ln (x+ \sqrt{1-x^2})}{x}dx,$ which becomes
$$ \int_{0}^{\frac {\pi}{2}} \frac {\ln \sin (t+ \frac {\pi}{4})}{\sin t}dt$$
It seems difficult to solve the particular integral. Any help?
 [/INST] First we establish that $$\int_0^{1}\frac{\ln(x+\sqrt{1-x^2})}{x}dx=\frac{\sqrt{\pi}}{4}\color{green}{\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}\frac{\Gamma\left(\frac{n}{2}\right)}{\Gamma\left(\frac{n+1}{2}\right)}}$$
So we proceed by substituting $x=\cos y$, then we have $$\int_0^{\frac{\pi}{2}}\frac{\ln\left(x+\sqrt{1-x^2}\right)}{x}dx=\int_0^{\frac{\pi}{2}}\frac{\ln\left(\cos y+\sin y\right)}{\cos y}\sin ydy$$  make the change of variable $y\mapsto \frac{\pi}{2} -y$ and hence of adding the obtained integral with latter integral, we get
$$\int_0^{\frac{\pi}{2}}\frac{\ln\left(\cos y+\sin y\right)}{\cos y}\sin ydy=\frac{1}{2}\int_0^{\frac{\pi}{2}}\frac{\ln\left(1+\sin y\right)}{\sin y}dy$$ For $ 0 < y< \frac{\pi}{2}$ , $ 0< \sin y <1$ and hence last integral reduces to( by series expansion )
$$\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{2n}\int_0^{\frac{\pi}{2}}\sin ^{n-1} y dy\overset{\text{Wallis' Int}}{=}\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{4n}\frac{\Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{n}{2}\right)}{\Gamma\left(\frac{1}{2}+\frac{n}{2}\right)}\\ \hspace{4.75cm}=\frac{\sqrt{\pi}}{4}{\color{green}{\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}\frac{\Gamma\left(\frac{n}{2}\right)}{\Gamma\left(\frac{n+1}{2}\right)}}}$$  Now split the sum into even and odd parity, giving us $$\sum_{n=1}^{\infty}\frac{\Gamma\left(n-\frac{1}{2}\right)}{(2n-1)\Gamma(n)}-\sum_{n=1}^{\infty}\frac{\Gamma(n)}{2n\Gamma\left(n+\frac{1}{2}\right)}$$
Now, we evaluate the former sum  $$\sum_{n=1}^{\infty}\frac{\Gamma\left(n-\frac{1}{2}\right)}{(2n-1)\Gamma(n)}=\sum_{n=0}^{\infty}\frac{\Gamma\left(n+\frac{1}{2}\right)}{(2n+1)\Gamma(n+1)}=\sqrt{\pi}{\color{red}{\sum_{n=0}^{\infty}\frac{1}{(2n+1)4^n}{2n\choose n}}}=\pi^{1/2}{\color{red}{\left[\frac{\pi}{2}\right]}}$$
By generating function of central binomial coefficients we $\displaystyle \sum_{n\geq 0}\frac{x^n}{4^n}{2n\choose n}=\frac{1}{\sqrt{1-x}}$. Replacing $x$ by $x^2$ on integrating from $ 0$ to $1$ gives $${\color{red}{\sum_{n=0}^{\infty}\frac{1}{(2n+1)4^n}{2n\choose n}}} =\int_0^1\frac{dx}{\sqrt{1-x^2}}={\color{red}{\frac{\pi}{2}}}$$ Further, we evaluate the latter sum $$\sum_{n=1}^{\infty}\frac{\Gamma(n)}{2n\Gamma\left(n+\frac{1}{2}\right)}=\frac{\pi^{-1/2}}{2}\color{blue}{\sum_{n=1}^{\infty}\frac{4^n}{n^2}{2n\choose n}^{-1}}=\frac{\pi^{-1/2}}{2}\color{blue}{\left[\frac{\pi^2}{2}\right]}$$ we use the generating function $\displaystyle \arcsin^2(x)=\frac{1}{2}\sum_{n=1}^{\infty}\frac{(2x)^{2n}}{n^2{2n\choose n}}$ (see here). Set $x=1$ we have our answer $\frac{\pi^2}{4}$.
Therefore our final answer for the series
$$\sum_{n\geq 1}\frac{(-1)^{n+1}}{n}\frac{\Gamma\left(\frac{n}{2}\right)}{\Gamma\left(\frac{n+1}{2}\right)}=\frac{\pi^{3/2}}{2}-\frac{\pi^{3/2}}{4}=\frac{\pi^{3/2}}{4}$$ On multiply by $\frac{\pi^{1/2}}{4}$ gives us
$$\int_0^1\frac{\ln\left(x+\sqrt{1-x^2}\right)}{x}dx=\frac{\pi^2}{16}$$

If one  is curious to derive the cited generating function then it quite easy to see by Lehmer identity, AMM, 1985 $$\sum_{m\geq 1}\frac{(2x)^{2m}}{m{2m\choose m}}=\frac{2x\arcsin(x)}{\sqrt{1-x^2}}$$ Dividing by by $x$ and hence on integrating gives us the desired form. For more interesting series on central binomial coefficients due to Lehmer, see here.

If we replace $-x^2$ by $x^2$, then the following equality holds.
$$\int_0^1\frac{\log\left(x+\sqrt{1+x^2}\right)}{x}dx=\frac{\pi^2}{12}-\frac{\operatorname{Li}_2\left(3-2\sqrt 2\right)}{2}-\frac{\log^2(1+\sqrt 2)}{2}+\log(2)\log(1+\sqrt 2)$$
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