Some Log integrals related to Gamma value Two years ago I evaluated some integrals related to $\Gamma(1/4)$.
First example:
$$(1)\hspace{.2cm}\int_{0}^{1}\frac{\sqrt{x}\log{(1+\sqrt{1+x})}}{\sqrt{1-x^2}} dx=\pi-\frac{\sqrt {2}\pi^{5/2}+4\sqrt{2}\pi^{3/2}}{2\Gamma{(1/4)^{2}}}.$$
The proof I have is based on the following formula concerning the elliptic integral of first kind (integrating both sides with carefully).
$$i \cdot K(\sqrt{\frac{2k}{1+k}})=K(\sqrt{\frac{1-k}{1+k}})-K(\sqrt{\frac{1+k}{1-k}})\cdot\sqrt{\frac{1+k}{1-k}}$$
for $0<k<1$.
\begin{align}
(2)\hspace{.2cm}\int_{0}^{1}\frac{\sqrt{2x-1}-2x  \arctan{(\sqrt{2x-1})}}{\sqrt{x(1-x)}(2x-1)^{3/2}}dx=\frac{\sqrt{2}\pi^{5/2}}{\Gamma{(1/4)}^2}-\frac{\sqrt{2\pi}\Gamma{(1/4)}^2}{8}.
\end{align}
\begin{align}
(3)\hspace{.2cm}\int_{0}^{\pi/2}\frac{\sin{x}\log{(\tan{(x/2))}+x}}{\sqrt{\sin{x}}(\sin{x}+1)}dx=\pi-\frac{\sqrt{2\pi}\Gamma{(1/4)}^{2}}{16}-\frac{\sqrt{2}\pi^{5/2}}{2\Gamma{(1/4)}^{2}}.
\end{align}
Could you find a solution to (2) and (3) employing only Beta function or other method?
I've tried with Mathematica, Mapple, etc and seems that this evaluations are not so well known.
Question is an improvement of (1) that has been proved in an elementary approach.
 A: This might be helpful (although not yet a solution). Define
$$I_\pm=\int_{0}^{1}\frac{\sqrt{x}\log{(\sqrt{1+x}\pm 1)}}{\sqrt{1-x^2}} \,dx.$$
Then $I_+ + I_-$ reduces to a Beta function integral,
$$I_+ + I_-=\int_{0}^{1}\frac{\sqrt{x}\log x}{\sqrt{1-x^2}} \,dx=(\pi -4) \sqrt{\pi }\,\frac{ \Gamma (3/4)}{\Gamma (1/4)}.$$
A: I also played around with this integral. My solution is a bit shorter than the OPs: First use the trick by @Claude and define
$$\tag{1}
I(a)=\int_0^1 \mathrm dx \sqrt\frac{x}{1-x^2}\log(a+\sqrt{1+x}),
$$
such that
$$\tag{2}
I(1) = I(0) + \int_0^1 \mathrm da \, I'(a).
$$
Partial fraction decomposition of $I'(a)$ gives
\begin{align}
I'(a) &= \int_0^1 \mathrm dx \frac{\sqrt\frac{x}{1-x^2}}{a+\sqrt{1+x}} 
\,\frac{a-\sqrt{1+x}}{a-\sqrt{1+x}}\tag{3a}\\
&=\int_0^1 \mathrm dx \frac{\sqrt\frac{x}{1-x}}{1+x-a^2}
+ \int_0^1 \mathrm dx \frac{a\sqrt\frac{x}{1-x^2}}{a^2-x-1}\tag{3b}.
\end{align}
Now we exchange the integration order in the second term only and also move $I(0)$ into the second term, to get the result
\begin{align}
I(1) &= 
\int_0^1 \mathrm da \int_0^1 \mathrm dx \frac{\sqrt\frac{x}{1-x}}{1+x-a^2} 
&+&
\int_0^1 \mathrm dx \int_0^1 \mathrm da \frac{a\sqrt\frac{x}{1-x^2}}{a^2-x-1}+I(0) \tag{4a}\\
&= \int_0^1 \mathrm da \, \pi\left(1-\sqrt{\tfrac{1-a^2}{2-a^2}}\right)
&+&
\int_0^1 \mathrm dx \sqrt{\tfrac{x}{1-x^2}} \log\sqrt x\tag{4b}\\
&= \pi-\pi^{3/2}\frac{\Gamma\left(\tfrac{3}{4}\right)}{\Gamma\left(\tfrac{1}{4}\right)} 
&+&\,\,
\frac{(\pi-4)\sqrt\pi \,\Gamma\left(\tfrac{3}{4}\right) }{2\Gamma\left(\tfrac{1}{4}\right)}\tag{4c}\\
&=\pi-\frac{(\pi+4)\sqrt\pi \,\Gamma\left(\tfrac{3}{4}\right) }{2\Gamma\left(\tfrac{1}{4}\right)}.\tag{4d}
\end{align}
Note that in (4c) the Beta function integral as motivated by @Carlo was used.
A: I was hoping to finish it but I am stuuck with a last integral. Then, this is just a comment.
Consider
$$I(a)=\int_{0}^{1}\frac{\sqrt{x}\log{(a+\sqrt{1+x})}}{\sqrt{1-x^2}}\, dx$$
$$I'(a)=\int_{0}^{1} \frac{\sqrt{x}}{\sqrt{1-x^2} \left(a+\sqrt{x+1}\right)}\,dx$$
$$I'(a)=\pi-\pi  \sqrt{1+\frac{1}{a^2-2}}-2 a K(-1)+2 a\, \Pi
   \left(\left.\frac{1}{a^2-1}\right|-1\right)$$
$$I(1)=\pi-K(-1) -\sqrt{\frac{\pi }{2}} \Gamma \left(\frac{3}{4}\right)^2+2\color{red}{\int_0^1  a\, \Pi
   \left(\left.\frac{1}{a^2-1}\right|-1\right)\,da}$$
$$I(0)=\frac 12\int_0^1 \frac{\sqrt{x} \log (x+1)}{\sqrt{1-x^2}}\,dx$$
$$I(0)=-\sqrt{\pi }
   (\gamma +\log (2))\frac{ \Gamma \left(\frac{3}{4}\right)}{\Gamma
   \left(\frac{1}{4}\right)}+$$ $$\frac{\pi ^{3/2}
  }{4 \sqrt{2}} \text{HypergeometricPFQRegularized}^{(\{0,0,0\},\{0,1\},0)}\left(
   \left\{\frac{1}{2},\frac{1}{2},\frac{1}{2}\right\},\left\{2,\frac
   {1}{2}\right\},\frac{1}{2}\right)$$
