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.
 A: A (very elegant) solution by Cornel Ioan Valean
Let's return first to a  result from the book, (Almost) Impossible Integrals, Sums, and Series, more precisely to the following very useful result, $\displaystyle 2\int_0^{\infty}\frac{t\log(x)}{(x+1)^2+t^2}\textrm{d}x=\arctan(t)\log(1+t^2)$, (see page $152$, eq. $3.149$) which is very easily proved by exploting the elementary result used in the same book, that is $\displaystyle \int_0^{\infty} \frac{\log(x)}{(x+a)(x+b)}\textrm{d}x=\frac{1}{2}\left(\frac{\log ^2(a)-\log^2(b)}{a-b}\right), \ a,b>0$, (see page $152$, eq. $3.150$) where using the symmetry is enough to get a proof. Finally,  we set $a=1+i t$ and $b=1-i t$.
Now, let's return to the main integral where we let $t\mapsto t^2$, and then we have
$$\mathcal{I}=2\int_0^{\infty} \frac{t\arctan(t)\log{(1+t^2)}}{4+t^4} \textrm{d}t=4\int_0^{\infty} \frac{t}{4+t^4}\left( \int_0^{\infty}\frac{t\log(x)}{(x+1)^2+t^2}\textrm{d}x\right)\textrm{d}t$$
$$=4\int_0^{\infty}\left( \int_0^{\infty}\frac{t^2\log(x)}{((x+1)^2+t^2)(4+t^4)}\textrm{d}t\right)\textrm{d}x=\pi\int_0^{\infty } \frac{  \log (x)}{x^2+4 x+5} \textrm{d}x. \tag1$$
Next, if we let $x\mapsto 5x$ in the last integral, we get
$$\mathcal{I}=\pi\int_0^{\infty} \frac{  \log (x)}{x^2+4 x+5} \textrm{d}x=\pi\int_0^{\infty}\frac{\log(5 x)}{5x^2+4 x+1} \textrm{d}x=\log(5)\pi\int_0^{\infty}\frac{1}{5x^2+4 x+1} \textrm{d}x$$
$$+\pi\underbrace{\int_0^{\infty}\frac{\log(x)}{5x^2+4 x+1} \textrm{d}x}_{\text{Next we let} \  \displaystyle x\mapsto 1/x}=\log(5)\arctan\left(\frac{1}{2}\right)\pi-\mathcal{I},$$
whence the desired result is obtained

$$\mathcal{I}=\frac{1}{2}\log(5)\arctan\left(\frac{1}{2}\right)\pi.$$

A first note: This simple strategy can also help for getting generalizations, and at the same time one can obtain a lot of other interesting results by using the main auxiliary result stated above. A very nice such example may be met in the book, (Almost) Impossible Integrals, Sums, and Series, in particular in Sect. 3.26, pages $150$-$154$.
A second note: Another interesting solution could be built by considering the parametrized integral, $\displaystyle \mathcal{I(a,b)}=\int_0^{\infty} \frac{t\arctan(a t)\log{(1+b^2 t^2)}}{4+t^4} \textrm{d}t$, where one then uses the differentiation with respect to both $a$ and $b$.
A: 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,
$$I=2\pi\int\limits_0^\infty \dfrac{\ln z}{(z+2+i)(z+2-i)}\text{ d}z 
= 2\pi\dfrac{\ln^2(2+i)-\ln^2(2-i)}{4i} = 2\pi\dfrac{(\ln5+i\operatorname{arccot} 2)^2-(\ln5-i\operatorname{arccot}2)^2}{4i}
 = \color{brown}{\mathbf{\dfrac\pi2\, \arctan\left(\frac12\right)\,\ln5}}.$$
A: For $a>0$,
$$\begin{aligned}I = \int_0^\infty  {\frac{{\log (1 + x)\arctan \sqrt x }}{{{a^4} + {x^2}}}dx}  &= \int_{ - \infty }^\infty  {\frac{x}{{{a^4} + {x^4}}}\log (1 + {x^2})\arctan xdx} \\ &= -\Im \int_{ - \infty }^\infty  {\frac{x}{{{a^4} + {x^4}}}{{\log }^2}(1 - ix)dx} \end{aligned}$$
The integrand is holomorphic on upper half plane, integral around big semicircle tends to $0$, calculating residues at $a\zeta, a\zeta^3$ (with $\zeta = e^{\pi i /4}$) give $$ I= \frac{{ \pi }}{{2{a^2}}}\Im\left[ {{{\log }^2}(1 + a\zeta ) - {{\log }^2}(1 - a{\zeta ^3})} \right]$$
when $a=\sqrt{2}$, it becomes $\frac{1}{2}\pi\arctan(1/2)\log 5$.
