The question is to evaluate $$\int_0^{\infty} \frac{\tan^{-1} x}{x(1+x^2)} dx$$
I used the substitution $x=\tan a$ then the given integral becomes $\int_0^{\pi/2} \frac{\tan^{-1}(\tan a)}{\tan a} da$ Now $\tan^{-1} (\tan a)=a \forall a \in [0,\pi /2]$ so that the integrand becomes $a/ \tan a$.i am facing trouble evaluating this.i tried using $a \to \pi/2 -a$ but couldn't simplify.Any ideas?
More generally, let $$J(t) = \int_0^\infty \frac{\tan^{-1}(tx)}{x(1+x^2)}\; dx$$
Of course $J(0)=0$, while (differentiating under the integral sign)
$$ J'(t) = \int_0^\infty \frac{dx}{(1+t^2 x^2)(1+x^2)}$$
which can be done by partial fractions, obtaining
$$ J'(t) = \frac{t \pi}{2(t^2-1)} - \frac{\pi}{2(t^2-1)} = \frac{\pi}{2(t+1)} $$
Now integrate that from $t=0$ to $1$.
The same integral can be found here Solve this integral:$\int_0^\infty\frac{\arctan x}{x(x^2+1)}\mathrm dx$
One more evaluation related with your approach. By letting $t=\arctan x$, then $dt=\dfrac{dx}{1+x^2}$ and $$ \begin{align} \int_0^{+\infty}\frac{\arctan x}{x(1+x^2)}dx&=\int_0^{\pi/2} t\frac{\cos t}{\sin t}dt=\left[t\log(\sin t)\right]_0^{\pi/2}-\int_0^{\pi/2} \log(\sin t)dt\\&=0-\left(-\frac{\pi\log 2}{2}\right)=\frac{\pi\log 2}{2} \end{align} $$ where we used another know result: Evaluate $\int_0^{\pi/2}\log\cos(x)\,\mathrm{d}x$
There is a hidden symmetry: $$\begin{eqnarray*} \int_{0}^{+\infty}\frac{\arctan x}{x(1+x^2)}\,dx &\stackrel{x\mapsto \tan\theta}{=}&\int_{0}^{\pi/2}\frac{\theta\,d\theta}{\tan\theta}\stackrel{\text{IBP}}{=}-\int_{0}^{\pi/2}\log\sin(\theta)\,d\theta\\&=&-\frac{1}{2}\int_{0}^{\pi}\log(\sin\theta)\,d\theta\tag{A}\end{eqnarray*} $$ and due to the trigonometric identity $$ \prod_{k=1}^{n-1}\sin\left(\frac{\pi k}{n}\right) = \frac{2n}{2^n}\tag{B} $$ the RHS of $(A)$ can be computed as the limit of a Riemann sum: $$\int_{0}^{\pi}\log\sin(\theta)\,d\theta = \lim_{n\to +\infty}\frac{\pi}{n}\log\prod_{k=1}^{n-1}\sin\left(\frac{\pi k}{n}\right)=\lim_{n\to +\infty}\frac{\pi\left(\log 2+\log n-n\log 2\right)}{n} $$ so that: $$ \int_{0}^{+\infty}\frac{\arctan x}{x(1+x^2)}\,dx = \color{red}{\frac{\pi\log 2}{2}}.\tag{C}$$
Proceeding by your method:
$\int_0^{\pi/2}\cfrac{a}{\tan(a)}da=\int_0^{\pi/2}a \tan(a)da$ [Now use Integration by Parts]
$= [a\log(\sin(a))]_0^{\pi/2} - \int_0^{\pi/2}\log\sin(a)da$
$= 0 - \int_0^{\pi/2}\log\sin(a)da$
$= \cfrac{\pi\ln(2)}{2}$
Instead of using Feynman's trick and differentiating under the integral sign as @Robert Israel did, in a very similar manner one may evaluate the integral by converting it to a double integral first.
Noting that $$\int^1_0 \frac{dt}{1 + x^2 t^2} = \frac{\tan^{-1} x}{x},$$ the integral may be rewritten as $$I = \int^\infty_0 \frac{\tan^{-1} x}{x(1 + x^2)} \, dx = \int^\infty_0 \int^1_0 \frac{1}{(1 + x^2 t^2)(1 + x^2)} \, dt dx.$$ Changing the order of integration then integrating yields \begin{align*} I &= \int^1_0 \int^\infty_0 \frac{1}{(1 + x^2 t^2)(1 + x^2)} \, dx dt\\ &= \int^1_0 \int^\infty_0 \left [\frac{1}{1 - t^2} \cdot \frac{1}{1 + x^2} - \frac{t^2}{1 - t^2} \cdot \frac{1}{1 + t^2 x^2} \right ] \, dx dt\\ &= \frac{\pi}{2} \int^1_0 \left [\frac{1}{1 - t^2} - \frac{t}{1 - t^2} \right ] \, dt\\ &= \frac{\pi}{2} \int^1_0 \frac{dt}{1 + t}\\ &= \frac{\pi}{2} \ln (1 + t) \Big{|}^1_0\\ &= \frac{\pi}{2} \ln (2). \end{align*}
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \int_{0}^{\infty}{\arctan\pars{x} \over x\pars{1 + x^{2}}}\,\dd x & \,\,\,\stackrel{x\ =\ \tan\pars{\theta}}{=}\,\,\, \int_{0}^{\pi/2}{\theta \over \tan\pars{\theta}}\,\dd\theta \\[5mm] & = \left.\Re\int_{\theta = 0}^{\theta = \pi/2}{-\ic\ln\pars{z} \over \bracks{\pars{z - 1/z}/2\ic}/\bracks{\pars{z + 1/z}/2}}\,{\dd z \over \ic z} \right\vert_{\ds{\ z\ =\ \exp\pars{\ic\theta}}} \\[5mm] & = \left.-\,\Im\int_{\theta = 0}^{\theta = \pi/2}{\pars{1 + z^{2}}\ln\pars{z} \over 1 - z^{2}}\,{\dd z \over z} \right\vert_{\ds{\ z\ =\ \exp\pars{\ic\theta}}} \\[1cm] & \stackrel{\mrm{as}\ \epsilon\ \to\ 0^{+}}{\sim}\,\,\, \Im\int_{1}^{\epsilon}{\pars{1 - y^{2}}\bracks{\ln\pars{y} + \pi\ic/2} \over 1 + y^{2}}\,{\ic\,\dd y \over \ic y} \\[2mm] + &\ \Im\int_{\pi/2}^{0}\bracks{\ln\pars{\epsilon} + \ic\theta}\,{\epsilon\expo{\ic\theta}\ic\,\dd\theta \over \epsilon\expo{\ic\theta}} + \Im\int_{\epsilon}^{1}{\pars{1 + x^{2}}\ln\pars{x} \over 1 - x^{2}} \,{\dd x \over x} \\[1cm] & = -\,{\pi \over 2}\int_{\epsilon}^{1}{1 - y^{2} \over 1 + y^{2}}\,{\dd y \over y} - {\pi \over 2}\,\ln\pars{\epsilon} \\[5mm] & \stackrel{\mrm{as}\ \epsilon\ \to\ 0^{+}}{\to}\,\,\, -\,{\pi \over 2}\int_{0}^{1}\pars{{1 - y^{2} \over 1 + y^{2}} - 1} \,{\dd y \over y} = \pi\int_{0}^{1}{y \over 1 + y^{2}}\,\dd y \\[5mm] & = \bbx{{\pi \over 2}\,\ln\pars{2}} \approx 1.0888 \end{align}
