Prove: $\int_0^{\infty} \frac{\ln{(1+x)}\arctan{(\sqrt{x})}}{4+x^2} \, \mathrm{d}x = \frac{\pi}{2} \arctan{\left(\frac{1}{2}\right)} \ln{5}$ Prove: $$\int_0^{\infty} \frac{\ln{(1+x)}\arctan{(\sqrt{x})}}{4+x^2} \, \mathrm{d}x = \frac{\pi}{2} \arctan{\left(\frac{1}{2}\right)} \ln{5}$$
This might be a repeat question (I couldnt find a question of this here).  If im being honest I dont know the first step really...  Maybe a clever integration by parts, substitution, differentiation under integral sign, power series, or contour?  If someone could give advice.

Let us try to collect the most interesting ideas in one simple solution.
At first,
$$I=\int\limits_0^\infty \dfrac{\ln(1+x)\arctan\sqrt x}{x^2+4}\text{ d}x
=\int\limits_0^\infty \dfrac{\ln(1+y^2)\arctan y}{y^4+4}\,2y\text{ d}y.$$
At the second, by pisco,
$$\ln(1+y^2) = \ln(1+iy) + \ln(1-iy),\\
\arctan y = \dfrac i2(\ln(1-iy) - \ln(1+iy)),$$
$$\ln(1+y^2)\arctan y = \dfrac i2(\ln^2(1-iy)-\ln^2(1+iy)).$$
Therefore,
$$I=\int\limits_{0}^\infty \dfrac{\ln^2(1+iy)-\ln^2(1-iy)}{4iy}\dfrac {8y^2\text{ d}y}{y^4+4}.\tag1$$
Taking in account Sophie Germain identity
$$y^4+4 = (y^2+2)^2 - 4y^2 = (y^2-2y+2)(y^2+2y+2),$$
easily to get
$$\int\limits_0^\infty \dfrac{y^2}{y^4+4}\text{ d}y=\dfrac\pi4,\quad
\int\limits_0^\infty \dfrac{1}{y^4+4}\text{ d}y=\dfrac\pi8,\quad
\int\limits_0^\infty \dfrac{1}{y^2+z^2}\text{ d}y=\dfrac\pi{2z}\tag2$$
(see also Wolfram Alpha integral1, integral2, integral3).
Now, applying the definite integral
$$\int\limits_0^\infty \dfrac{\ln t}{(t+a)(t+b)} = \dfrac{\ln^2a - \ln^2 b}{2a-2b},\tag3$$
which is known from the answer of user9735739 (it should be correct if $\Re a >0,\ \Re b >0$), in the form of
$$\int\limits_0^\infty\dfrac{\ln z\text{ d}z}{(z+1+iy)(z+1-iy)} = \dfrac{\ln^2(1+iy)-\ln^2(1-iy)}{4iy},$$
integral $(1)$ can be presented in the form of
$$I=\int\limits_{0}^\infty \int\limits_0^\infty \dfrac{\ln z}{(z+1)^2+y^2}\dfrac {8y^2}{y^4+4}\text{ d}z\text{ d}y= 8 \int\limits_{0}^\infty J(z) \ln z \text{ d}z,\tag4$$
where
\begin{align}
&J(z-1) = \int\limits_0^\infty \dfrac{y^2\text{ d}y}{(y^2+z^2)(y^4+4)}
 = \dfrac1{z^4+4}\int\limits_0^\infty \left(\dfrac{z^2 y^2}{y^4+4}+\dfrac{4}{y^4+4}-\dfrac{z^2}{y^2+z^2}\right)\text{ d}y\\[4pt]
&=\dfrac1{z^4+4}\left(z^2\cdot\dfrac\pi4 + 4\cdot\dfrac\pi8 - z^2\cdot\dfrac\pi{2z}\right)
 = \dfrac\pi4\dfrac{z^2-2z+2}{z^4+4} = \dfrac\pi{4((z+1)^2+1)},\\[4pt]
&J(z) = \dfrac\pi{4((z+2)^2+1)}.
\end{align}
Finally, applying $(3)$ once more,
$$I=2\pi\int\limits_0^\infty \dfrac{\ln z}{(z+2+i)(z+2-i)}\text{ d}z 
= 2\pi\dfrac{\ln^2(2+i)-\ln^2(2-i)}{4i} = 2\pi\dfrac{(\ln5+i\operatorname{arccot} 2)^2-(\ln5-i\operatorname{arccot}2)^2}{4i}
 = \color{brown}{\mathbf{\dfrac\pi2\, \arctan\left(\frac12\right)\,\ln5}}.$$
