I may as well give an answer:
Note $$ \frac{ e^{-at} - e^{-bt} }{t} = \int^b_a e^{-xt} dx $$ so our integral is
$$ \int^{\infty}_0 \int^b_a e^{-xt} dx dt = \int^b_a \int^{\infty}_0 e^{-xt} dt dx $$
$$ = \int^b_a \frac{1}{x} dx = \log(b/a) $$
This is a general method, and often this whole process is compressed into a well known integral called Frullani's Integral.
Starting with the following identity that can be found on page $95$ Eq $(5)$ of this paper
$$\sum_{n=1}^\infty \overline{H}_n\frac{x^n}{n}=\operatorname{Li}_2\left(\frac{1-x}{2}\right)-\operatorname{Li}_2(-x)-\ln2\ln(1-x)-\operatorname{Li}_2\left(\frac12\right)$$
Multiply both sides by $\large \frac{\ln(1-x^2)}{x}$ then integrate from $x=0$ to $x=1$ we get
$$\underbrace{\sum_{n=1}^\infty \frac{\overline{H}_n}{n}\int_0^1x^{n-1}\ln(1-x^2)\ dx}_{\large S}$$
$$\small{=I-\underbrace{\int_0^1\frac{\operatorname{Li}_2(-x)\ln(1-x^2)}{x}\ dx}_{\large J}-\ln2\underbrace{\int_0^1\frac{\ln(1-x)\ln(1-x^2)}{x}\ dx}_{\large K}-\operatorname{Li}_2\left(\frac12\right)\underbrace{\int_0^1\frac{\ln(1-x^2)}{x}\ dx}_{\large -\frac12\zeta(2)}}$$
or
$$I=S+J+\ln2\ K-\frac12\zeta(2)\operatorname{Li}_2\left(\frac12\right)\tag1$$
Evaluating $S$
Notice that
$$\int_0^1 x^{n-1}\ln(1-x^2)\ dx\overset{x^2\to x}{=}\frac12\int_0^1 x^{n/2-1}\ln(1-x)\ dx=-\frac{H_{n/2}}{n}$$
$$\Longrightarrow S=\boxed{-\sum_{n=1}^\infty\frac{\overline{H}_nH_{n/2}}{n^2}}$$
Evaluating $J$
Writing $\ln(1-x^2)=\ln(1-x)+\ln(1+x)$ gives
$$J=\int_0^1\frac{\operatorname{Li}_2(-x)\ln(1-x)}{x}dx+\int_0^1\frac{\operatorname{Li}_2(-x)\ln(1+x)}{x}dx$$
$$=\sum_{n=1}^\infty\frac{(-1)^n}{n^2}\int_0^1 x^{n-1}\ln(1-x)\ dx-\frac12\operatorname{Li}_2^2(-x)|_0^1$$
$$=-\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^3}-\frac{5}{16}\zeta(4)$$
$$=\boxed{-2\operatorname{Li_4}\left(\frac12\right)+\frac{39}{16}\zeta(4)-\frac74\ln2\zeta(3)+\frac12\ln^22\zeta(2)-\frac{1}{12}\ln^42}$$
where we used $\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^3}$$=2\operatorname{Li_4}\left(\frac12\right)-\frac{11}4\zeta(4)+\frac74\ln2\zeta(3)-\frac12\ln^22\zeta(2)+\frac{1}{12}\ln^42$
Evaluating $K$
Similarly
$$K=\underbrace{\int_0^1\frac{\ln^2(1-x)}{x}\ dx}_{\large 2\zeta(3)}+\underbrace{\int_0^1\frac{\ln(1-x)\ln(1+x)}{x}\ dx}_{\large -\frac{5}{8}\zeta(3)}=\boxed{\frac{11}8\zeta(3)}$$
where the second integral is evaluated here.
Plug the boxed results in $(1)$ we obtain that
$$I=-2\operatorname{Li}_4\left(\frac12\right)+\frac{29}{16}\zeta(4)-\frac38\ln2\zeta(3)+\frac34\ln^22\zeta(2)-\frac1{12}\ln^42-\sum_{n=1}^\infty\frac{\overline{H}_nH_{n/2}}{n^2}$$
In here we proved
$$\sum_{n=1}^\infty\frac{\overline{H}_nH_{n/2}}{n^2}=\frac1{24}\ln^42-\frac14\ln^22\zeta(2)+\frac{21}{8}\ln2\zeta(3)-\frac{9}{8}\zeta(4)+\operatorname{Li}_4\left(\frac12\right)$$
$$\Longrightarrow I=-\frac1{8}\ln^42+\ln^22\zeta(2)-3\ln2\zeta(3)+\frac{47}{16}\zeta(4)-3\operatorname{Li}_4\left(\frac12\right)$$
Best Answer
Since $\displaystyle \frac{\arctan{x}}{x} = \int_0^1 \frac{1}{1+x^2y^2}\, \mathrm{dy}$, we have:
$$\begin{aligned} I & = \int_0^{\infty} \frac{\arctan{x}}{x(1+x^2)}\,\mathrm{dx} \\& = \int_0^{\infty} \int_0^1 \frac{1}{(1+x^2)(1+x^2y^2)}\,\mathrm{dy}\,\mathrm{dx} \\& = \int_0^{1} \int_0^\infty \frac{1}{(1+x^2)(1+x^2y^2)}\,\mathrm{dx}\,\mathrm{dy} \\& =\frac{\pi}{2} \int_0^{1} \frac{1}{1+y}\,\mathrm{dy} \\& =\pi \log \sqrt{2}. \end{aligned} $$