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.
 A: Substitute $t=\sqrt x$ in $I= \displaystyle\int _0^1\frac{\ln (1+\sqrt{x})}{1+x^2}\:dx$
\begin{align}
I &
=2\int _0^1\frac{t \ln \left(1+t\right)}{1+t^4}dt
\overset{t\to\frac1t}= \int _0^\infty\frac{t \ln \left(1+t\right)}{1+t^4}dt - \int _1^\infty\overset{t^2\to t}{\frac{t \ln t}{1+t^4}}dt \\
&=  -\frac12\int _0^\infty{\ln \left(1+t\right)}\>d(\cot^{-1}t^2) -\frac14\int_1^\infty\frac{\ln t}{1+t^2}dt\\
&= \frac12 \int _0^\infty\frac{\cot^{-1}t^2}{1+t}dt - \frac14 G\tag1
\end{align}
Let $J(a) = \displaystyle \int _0^\infty\frac{\cot^{-1}(at^2)}{1+t}dt$ and differentiate under the integral
\begin{align}
J’(a) &= -\int _0^\infty\frac{t^2}{(1+t)(1+a^2t^4)}dt\\
&= \frac1{1+a^2} \int _0^\infty\left(-\frac1{1+t}+\frac{a^2t^3}{1+a^2t^4}-\frac t{1+ a^2t^4}+\frac{1-a^2t^2}{1+a^2t^4}\right)dt\\
&=\frac1{1+a^2} \left(\frac12\ln a -\frac\pi{4a}-\frac{\pi(a-1)}{2\sqrt{2a}}\right)
\end{align}
Then, with $J(\infty) =0$
\begin{align}
& \int _0^\infty\frac{\cot^{-1}t^2}{1+t}dt  = J(1)=- \int_1^\infty J’(a)da\\
=&-\frac12\int_1^\infty \frac{\ln a}{1+a^2}da +\frac\pi4 \int_1^\infty \frac{1}{a(1+a^2)}da
 +\frac\pi2\int_1^\infty \frac{a-1}{\sqrt {2a}(1+a^2)}da\\
=& -\frac12G +\frac\pi8\ln2 + \frac\pi2\ln(1+\sqrt2)
\end{align}
Plug into $(1)$ to obtain
$$I= \frac{\pi }{16}\ln 2+\frac{\pi }{4}\ln \left(1+\sqrt{2}\right)-\frac{1}{2}G$$
A: 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.
