How to evaluate $\int _0^1\frac{\ln \left(1+\sqrt{x}\right)}{1+x^2}\:dx$ How can I approach:
$$\int _0^1\frac{\ln \left(1+\sqrt{x}\right)}{1+x^2}\:dx$$
I tried the usual differentiation under the integral sign but it didn't work so well.
I also tried to rewrite it the following 2 ways:
$$\int _0^1\frac{\ln \left(1+\sqrt{x}\right)}{1+x^2}\:dx=\int _0^1\frac{\ln \left(\sqrt{x}+x\right)}{1+x^2}\:dx-\frac{1}{2}\underbrace{\int _0^1\frac{\ln \left(x\right)}{1+x^2}\:dx}_{-G}$$
$$\int _0^1\frac{\ln \left(1+\sqrt{x}\right)}{1+x^2}\:dx=\underbrace{\int _0^1\frac{\ln \left(1-x\right)}{1+x^2}\:dx}_{\frac{\pi }{8}\ln \left(2\right)-G}-\int _0^1\frac{\ln \left(1-\sqrt{x}\right)}{1+x^2}\:dx$$
Yet I'm still stuck with those other integrals, any help is appreciated.
Managed to numerically find that:
$$\int _0^1\frac{\ln \left(1+\sqrt{x}\right)}{1+x^2}\:dx=\frac{\pi }{16}\ln \left(2\right)+\frac{\pi }{4}\ln \left(1+\sqrt{2}\right)-\frac{1}{2}G$$
But I still dont know how to approach it.

I do not think that I should arrive to the result but here are my attempts.
Trying to work the antiderivative
$$\int \frac{\log \left(1+\sqrt{x}\right)}{1+x^2}\,dx=2\int \frac{ t \log (t+1)}{t^4+1}\,dt$$ Using partial fractions
$$\frac{ t }{t^4+1}=\frac{ t }{(t-a)(t-b)(t-c)(t-d)}$$ where $(a,b,c,d)$ are the roots of unity
$$\frac{ t }{t^4+1}=\frac A{t-a}+\frac B{t-b}+\frac C{t-c}+\frac D{t-d}$$
$$A=\frac{a}{(a-b) (a-c) (a-d)} \qquad \qquad B=\frac{b}{(b-a) (b-c) (b-d)}$$
$$C=\frac{c}{(c-a) (c-b) (c-d)} \qquad \qquad D=\frac{d}{(d-a) (d-b) (d-c)}$$
So, four integrals
$$I_k=\int \frac{\log(t+1)}{t-k}\,dt=\text{Li}_2\left(\frac{t+1}{k+1}\right)+\log (t+1) \log
   \left(1-\frac{t+1}{k+1}\right)$$
$$J_k=\int_0^1 \frac{\log(t+1)}{t-k}\,dt=\log (2) \log \left(\frac{k-1}{k+1}\right)+\text{Li}_2\left(\frac{2}{k+1}\right)-\text{Li}_2\left(\frac{1}{k+1}\right)$$ Now, starts the nightmare !
Since @Quanto provided a nice answer, I give up.
