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?
Split the integral at $\frac{1}{\sqrt{2}}$ and use the substitution $x = \sqrt{1-y^2}$ in the second part to obtain \begin{align} I &\equiv \int \limits_0^1 \frac{\ln(x+\sqrt{1-x^2})}{x} \, \mathrm{d} x = \int \limits_0^{\frac{1}{\sqrt{2}}} \frac{\ln(x+\sqrt{1-x^2})}{x} \, \mathrm{d} x + \int \limits_{\frac{1}{\sqrt{2}}}^1 \frac{\ln(x+\sqrt{1-x^2})}{x} \, \mathrm{d} x \\ &= \int \limits_0^{\frac{1}{\sqrt{2}}} \frac{\ln(x+\sqrt{1-x^2})}{x} \, \mathrm{d} x + \int \limits_0^{\frac{1}{\sqrt{2}}} \frac{y \ln(y+\sqrt{1-y^2})}{1-y^2} \, \mathrm{d} y = \int \limits_0^{\frac{1}{\sqrt{2}}} \frac{ \ln(x+\sqrt{1-x^2})}{x(1-x^2)} \, \mathrm{d} x \, . \end{align} Now let $x = \sin (t/2)$ to find \begin{align} I &= \frac{1}{2} \int \limits_0^{\frac{\pi}{2}} \frac{\ln \left(\sin\left(\frac{t}{2}\right) + \cos\left(\frac{t}{2}\right)\right)}{\sin\left(\frac{t}{2}\right)\cos\left(\frac{t}{2}\right)} \, \mathrm{d} t = \frac{1}{2} \int \limits_0^{\frac{\pi}{2}} \frac{\ln \left[\left(\sin\left(\frac{t}{2}\right) + \cos\left(\frac{t}{2}\right)\right)^2\right]}{2\sin\left(\frac{t}{2}\right)\cos\left(\frac{t}{2}\right)} \, \mathrm{d} t \\ &= \frac{1}{2} \int \limits_0^{\frac{\pi}{2}} \frac{\ln\left(1+2\sin\left(\frac{t}{2}\right)\cos\left(\frac{t}{2}\right)\right)}{2\sin\left(\frac{t}{2}\right)\cos\left(\frac{t}{2}\right)} \, \mathrm{d} t = \frac{1}{2} \int \limits_0^{\frac{\pi}{2}} \frac{\ln\left(1+\sin(t)\right)}{\sin(t)} \, \mathrm{d} t \\ &=\frac{1}{2} \int \limits_0^{\frac{\pi}{2}} \frac{\ln\left(1+\cos(t)\right)}{\cos(t)} \, \mathrm{d} t \, . \end{align} Define (idea from this question) $$ f(a) \equiv \frac{1}{2} \int \limits_0^{\frac{\pi}{2}} \frac{\ln\left(1+\cos(a)\cos(t)\right)}{\cos(t)} \, \mathrm{d} t $$ for $ a \in [0,\frac{\pi}{2}]$ and observe that $f(0)=I$ and $f(\frac{\pi}{2}) = 0$. Compute (using $\tan(\frac{t}{2}) = s$) \begin{align} f'(a) &= - \frac{\sin(a)}{2} \int \limits_0^{\frac{\pi}{2}} \frac{1}{1+\cos(a)\cos(t)} \, \mathrm{d} t = - \sin(a) \int \limits_0^1 \frac{\mathrm{d} s}{1+\cos(a) + (1-\cos(a))s^2} \\ &= - \frac{\sin(a)}{1+\cos(a)} \sqrt{\frac{1+\cos(a)}{1-\cos(a)}} \arctan \left(\sqrt{\frac{1-\cos(a)}{1+\cos(a)}}\right) \\ &= - \frac{\sin(a)}{\sqrt{1-\cos^2 (a)}} \arctan\left(\tan\left(\frac{a}{2}\right)\right) = - \frac{a}{2} \, . \end{align} And finally, $$ I = f(0) = f \left(\frac{\pi}{2}\right) + \int \limits_{\frac{\pi}{2}}^0 f'(a) \, \mathrm{d} a = 0 + \int \limits_0^{\frac{\pi}{2}} \frac{a}{2} \, \mathrm{d} a = \frac{\pi^2}{16} \, .$$
Version 1. $$\int_{0}^{1}\frac{\log(x+\sqrt{1-x^2})}{x}\,dx = \int_{0}^{\pi/2}\log(\sin\theta+\cos\theta)\cot(\theta)\,d\theta \tag{1}$$ by enforcing the substitution $\theta\to\frac{\pi}{2}-\theta$ and averaging turns out to be equivalent to $$ \int_{0}^{\pi/2}\frac{\log(\sin\theta+\cos\theta)}{2\sin\theta\cos\theta}\,d\theta\stackrel{\theta\mapsto 2\arctan u}{=}\int_{0}^{1}\frac{\log(1+2t-t^2)-\log(1+t^2)}{2t(1-t^4)}\,dt\tag{2} $$ which can be managed by partial fraction decomposition, through the dilogarithm functional identities $(3)-(7)$, since $$ \int\frac{\log(1-t)}{t}\,dt = C-\text{Li}_2(t).\tag{3} $$ The same applies is we avoid the initial symmetrization, since $$ \int_{0}^{\pi/2}\frac{\log(\sin\theta+\cos\theta)}{\tan\theta}\,d\theta=\int_{0}^{1}\left[\log(1+2t-t^2)-\log(1+t^2)\right]\frac{1-t^2}{t(1+t^2)}\,dt .\tag{4}$$
Version 2. By immediately substituting $\theta=\arctan u$ in $(1)$, the original integral is converted into $$ \int_{0}^{+\infty}\frac{\log(1+u)-\frac{1}{2}\log(1+u^2)}{u(1+u^2)}\,du $$ which by Feynman's trick equals $$ \int_{0}^{1}\frac{\pi+2a\log a}{2(1+a^2)}\,da +\frac{1}{4}\int_{0}^{1}\frac{\log a}{1-a}\,da=\frac{\pi^2}{8}-\frac{\pi^2}{48}-\frac{\pi^2}{24}=\color{blue}{\frac{\pi^2}{16}}.\tag{5}$$ (Poly)logarithmic integrals always are a tricky thing, one never knows in advance what is the best moment for enforcing a substitution or exploiting some symmetry. In this case the usual tangent half-angle substitution just introduces a detour in a straightforward solution.
Version 3. By considering the Fourier series of $\log\sin$ and $\log\cos$ we have that, in a distributional sense related to $L^2(-\pi/2,\pi/2)$, $$ \cot\theta = 2 \sum_{k\geq 1} \sin(2k\theta) $$ $$ \log(\sin\theta+\cos\theta)=-\frac{\log 2}{2}-\sum_{k\geq 1}\frac{\cos(2k\theta+k\pi/2)}{k} $$ hence by Parseval's theorem $$ \int_{0}^{\pi/2}\log(\sin\theta+\cos\theta)\cot(\theta)\,d\theta =\frac{\pi}{4}\sum_{\substack{k\geq 1\\k\text{ odd}}}\frac{(-1)^{(k-1)/2}}{k}=\frac{\pi}{4}\cdot\frac{\pi}{4} = \color{red}{\frac{\pi^2}{16}}\tag{6}$$ ... WOW! This approach allows a simple and explicit evaluation of many integrals of the form $\int_{0}^{\pi/2}\log(\sin\theta+\cos\theta)\,\omega(\theta)\,d\theta$, thus many integrals of the form $\int_{0}^{1}\log(x+\sqrt{1-x^2})\,w(x)\,dx$. "Thinking backwards", the original problem can be probably tackled also by computing the moments $\int_{0}^{1}x^{2m+1} \log(x+\sqrt{1-x^2})\,dx$, then performing an interpolation/analytic continuation.
Let $I(a)=\int_0^1 \frac{\ln(a x+\sqrt{1-x^2})}{x}dx$. Then, $I(0) = \int_0^1 \frac{\ln\sqrt{1-x^2}}{x}dx \overset{x^2\to x} =-\frac{\pi^2}{24}$
$$I’(a)=\int_0^1 \frac{dx}{a x+\sqrt{1-x^2}} =\frac1{1+a^2}\left(\frac\pi2+a\ln a \right)$$
and
$$\int_0^1 \frac{\ln(x+\sqrt{1-x^2})}{x}dx =I(1)= I(0)+\int_0^1I’(a)da \\ \hspace{20mm}= -\frac{\pi^2}{24}+ \frac\pi2\int_0^1 \frac{da}{1+a^2} + \int_0^1 \frac{a\ln a}{1+a^2}da= \frac{\pi^2}{16} $$
Here is a way to calculate the integral without Feynman's trick or trigonometric substitutions.
By using the substitution previously used by Shashi we have: $$\underbrace{\int _0^1\frac{\ln \left(x+\sqrt{1-x^2}\right)}{x}\:dx}_{x=\frac{1}{\sqrt{1+t^2}}}=\underbrace{\int _0^{\infty }\frac{t\ln \left(\frac{1+t}{\sqrt{1+t^2}}\right)}{1+t^2}\:dt}_{t=\frac{1}{t}}$$ $$\int _0^{\infty }\frac{t\ln \left(\frac{1+t}{\sqrt{1+t^2}}\right)}{1+t^2}\:dt=\int _0^{\infty }\frac{\ln \left(\frac{1+t}{\sqrt{1+t^2}}\right)}{t}\:dt-\int _0^{\infty \:}\frac{t\ln \left(\frac{1+t}{\sqrt{1+t^2}}\right)}{1+t^2}\:dt$$ $$\int _0^{\infty }\frac{t\ln \left(\frac{1+t}{\sqrt{1+t^2}}\right)}{1+t^2}\:dt=\frac{1}{2}\int _0^{\infty }\frac{\ln \left(\frac{1+t}{\sqrt{1+t^2}}\right)}{t}\:dt$$ $$=\frac{1}{2}\int _0^1\frac{\ln \left(\frac{1+t}{\sqrt{1+t^2}}\right)}{t}\:dt+\frac{1}{2}\underbrace{\int _1^{\infty }\frac{\ln \left(\frac{1+t}{\sqrt{1+t^2}}\right)}{t}\:dt}_{t=\frac{1}{t}}=\int _0^1\frac{\ln \left(\frac{1+t}{\sqrt{1+t^2}}\right)}{t}\:dt$$ $$=\int _0^1\frac{\ln \left(1+t\right)}{t}\:dt-\frac{1}{2}\underbrace{\int _0^1\frac{\ln \left(1+t^2\right)}{t}\:dt}_{t=t^2}$$ $$=\frac{3}{4}\int _0^1\frac{\ln \left(1+t\right)}{t}\:dt=\frac{3}{4}\sum _{k=1}^{\infty }\frac{\left(-1\right)^{k+1}}{k^2}=\frac{3}{8}\zeta \left(2\right)$$ Thus: $$\int _0^1\frac{\ln \left(x+\sqrt{1-x^2}\right)}{x}\:dx=\frac{3}{8}\zeta \left(2\right)$$
As James Arathoon mentioned in the comments, by the substitution $t= \sqrt[]{\frac {1-x^2}{x^2}}$ the integral is equal to: \begin{align} I:=\int^1_0 \frac{\log(x+\sqrt[]{1-x^2} )}{x}\,dx=\int^\infty_0 \frac{t\log\left( \frac{t+1}{\sqrt[]{t^2+1}}\right)}{t^2+1}\,dt \end{align} One can rewrite it a bit: \begin{align} I=\frac 1 2 \int^\infty_0 \frac{t\log\left( \frac{(t+1)^2}{t^2+1}\right)}{t^2+1}\,dt = \frac 1 2 \int^\infty_0 \frac{t\log\left( 1+\frac{2t}{t^2+1}\right)}{t^2+1}\,dt \end{align} Now define the following function $F:[0, 1]\to\mathbb R$ as follows: \begin{align} F(a) := \frac 1 2 \int^\infty_0 \frac{t\log\left( 1+\frac{2at}{t^2+1}\right)}{t^2+1}\,dt \end{align} Using Feynman's Trick one gets: \begin{align} F'(a) = \int^\infty_0 \frac{t^2}{(t^2+1)(t^2+2at+1)}\,dt \end{align} This integral is not very hard to compute, for instance one can do it by partial fraction decomposition or contour integration to get: \begin{align} F'(a) =\frac{\arctan\left(\frac{\sqrt[]{1-a^2}}{a} \right)}{2\ \sqrt[]{1-a^2}} \end{align} We know that: \begin{align} I = F(1) = \int^1_0 F'(a)\,da = \frac{1}{2}\int^1_0 \frac{\arctan\left(\frac{\sqrt[]{1-a^2}}{a} \right)}{\sqrt[]{1-a^2}}\,da \end{align} This looks a bit scary, but hey it is very innocent after setting $a=\cos(x)$, because then one gets: \begin{align} I = \frac{1}{2}\int^0_{\pi/2} \frac{\arctan\left(\tan(x)\right)}{\sin(x)}(-\sin(x))\,dx = \frac{1}{2}\int^{\pi/2}_0 x\,dx = \frac{\pi^2}{16} \end{align}
Though there are already 6 wonderful solutions, I want to share mine with you now. Wish that you can enjoy it.
Letting $x=\cos \theta$ yields
$$ I=\int_{0}^{\frac{\pi}{2}} \frac{\ln (\cos \theta+\sin \theta)}{\cos \theta}\sin \theta d \theta $$
and letting $x=\sin \theta$ yields $$ I=\int_{0}^{\frac{\pi}{2}} \frac{\ln (\sin \theta+\cos \theta)}{\sin \theta} \cos \theta d \theta $$ Combining them gives $$ \begin{aligned} 2 I &=\int_{0}^{\frac{\pi}{2}} \frac{\ln (\sin \theta+\cos \theta)}{\sin \theta \cos \theta} d \theta \\ &=\int_{0}^{\frac{\pi}{2}} \frac{\ln (1+\sin 2 \theta)}{\sin 2 \theta} d \theta \\ &\stackrel{2\theta\mapsto\theta}{=} \frac{1}{2} \int_{0}^{\pi} \frac{\ln (1+\sin \theta)}{\sin \theta} d \theta \\ &=\int_{0}^{\frac{\pi}{2}} \frac{\ln (1+\sin \theta)}{\sin \theta} d \theta \quad \textrm{( By symmetry)}\\ &\stackrel{\theta\mapsto \frac{\pi}{2} -\theta}{=} \int_{0}^{\frac{\pi}{2}} \frac{\ln (1+\cos \theta)}{\cos \theta} d \theta \\ &=\frac{\pi^{2}}{8} \end{aligned} $$
Putting $a=0$ in my answer, we get $2 I=\dfrac{\pi^{2}}{8}$ and hence $\boxed{I=\frac{\pi^{2}}{16}}$.
:|D Wish you enjoy my solution!
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. $$\int_0^1\frac{\log\left(x+\sqrt{1+x^2}\right)}{x}dx=\frac{\pi^2}{12}-\frac{\operatorname{Li}_2\left(3-2\sqrt 2\right)}{2}-\frac{\log^2(1+\sqrt 2)}{2}+\log(2)\log(1+\sqrt 2)$$
