<s>[INST] 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.
 [/INST] 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,
$$I=2\pi\int\limits_0^\infty \dfrac{\ln z}{(z+2+i)(z+2-i)}\text{ d}z 
= 2\pi\dfrac{\ln^2(2+i)-\ln^2(2-i)}{4i} = 2\pi\dfrac{(\ln5+i\operatorname{arccot} 2)^2-(\ln5-i\operatorname{arccot}2)^2}{4i}
 = \color{brown}{\mathbf{\dfrac\pi2\, \arctan\left(\frac12\right)\,\ln5}}.$$
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